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ENGINEERING MECHANICS 



ENGINEERING MECHANICS 



A REVISION OF 

"NOTES ON MACHINE DESIGN" 

PREPARED BY OFFICERS OF THE 

DEPARTMENT OF MARINE ENGINEERING AND NAVAL CONSTRUCTION 
U. S. NAVAL ACADEMY 

COMBINED WITH 

THE MATHEMATICS AND GENERAL PRINCIPLES 

NECESSARY FOR THE SOLUTION OF 

THE PROBLEMS 

BV 

C. N. OFFLEY, U. S. N. 



DEPARTMENT OF MARINE ENGINEERING 

AND NAVAL CONSTRUCTION, 

U. S. NAVAL ACADEMY 



ANNAPOLIS, MD. 

THE UNITED STATES NAVAL INSTITUTE 

1911 






COPTKIGHT, 1911 
BY 

The United States Naval Institute 



HALTIMORE, MD., C. S. A. 



im 



CClA28tM08 



PREFACE 

The object of the course in engineering mechanics is the study of 
the theoretical and mathematical principles involved in the applica- 
tion of mechanics to various engineering problems, and then the 
application of these principles to the solution of actual problems. 
With this object in view, the problems are selected to bring out as 
many principles as possible. A number of recitations are first 
assigned, leading up to each problem, which are then followed by 
several practical work periods in the drafting room^ where a practical 
solution is worked out. Certain features of the theoretical study of 
strength of material are reviewed under the present subject. In 
particular it is the aim to point out the way in which many of the 
purely theoretical principles are modified in the solution of practical 
engineering problems. 

In the preparation of the following pages many standard works 
have been consulted and their methods adopted where it has seemed 
advisable. Among these are : " The Elements of Machine Design,^^ 
by Dr. W. C. Unwin ; " Elements of Machine Design/^ by Kimball 
and Barr ; " Strength of Materials,^^ by Professor Morley ; ^' Machine 
Design, Construction and Drawing," by Prof. H. J. Spooner: 
"Mechanics Applied to Engineering," by Prof. John G-oodman. 

The practical work problems are taken from " Notes on the De- 
sign of Propelling Machinery ' for N'aval Vessels," 1902. These 
notes are largely the work of Eear- Admiral J. K. Barton, U. S. ¥., 
when on duty as an instructor at the Naval Academy from 1890 to 
1893. The article on " Eiveting " is the work of Captain F. J. 
Schell, when at the Academy from 1891 to 1895. The article on the 
" Theory of the Connecting Eod and Problem IV," in its present 
form, is the work of Prof. Theo. W. Johnson. Others who con- 
tributed to these notes are Captains Bartlett, Kinkaid, Gow and 
Lieut.-Commander Moody. This collection of notes was revised and 
published in 1908 as " Notes on Machine Design," by Captain 
P. W. Bartlett, U. S. N., Head of the Department of Marine 
Engineering and Naval Construction. They have been thoroughly 
revised and taken bodily into the present work. It is hoped that the 



b PREFACE. 

features formerly found to be obscure have been simplified and 
cleared up. 

In all cases the endeavor has been made to adopts as far as pos- 
sible, the especial methods of the Bureau of Steam Engineering of 
the Navy Department. 

In the preparation of the book Captain P. W. Bartlett^ U. S. Navy, 
and Lieut.- Commander A. W. Hinds, U. S. Navy, have given in- 
valuable assistance in consultations, in checking the various mathe- 
matical formulge and the solutions to the Practical Problems; they 
have also read and corrected all the proof sheets. 

C. N. Oefley, U. S. N. 
Navy Yard, Puget Sound, October, 1910. 



CONTENTS 



CHAPTER I. 
Units, Definitions, Quality of Materials. — Forces Acting on a 
Machine 9 

CHAPTER 11. 
General Problem of Design. — Effect of Temperature. — Steady, Sud- 
den and Impulsive Loads. — Fatigue. — Woliler's Law. — Factors 
of Safety 17 

CHAPTER IIL 

Resilience, Sudden and Impulsive Loads. — Tables of Strength, etc. 
— Tension. — Compression. — Shearing. — Modulus of Rigidity.... 29 

CHAPTER IV. 
Bending. — Moments of Inertia. — Modulus of Section. — Table of I, 
Z, etc. — Tables and Curves oi Shearing Face, Bending Moments, 
Deflection 46 

CHAPTER V. 
Directions for Practical Work. — Practical Problem I. — The Knuckle 

Joint 68 

CHAPTER VI. 
' Bolts, Nuts, Screws. 

Forms of Screw Threads. Applications of Bolts, Studs, Nuts, etc. 
Stresses in Bolts and Screws. Friction and Efficiency of Screws. 
Number, Size and Pitch of Cylinder Cover Studs 76 

CHAPTER VII. 
Practical Problem II: The Screw Jack 99 

CHAPTER VIII. 
Compound Stresses: Bending Combined with Tension or Compres- 
sion. Combined Stresses. Struts and Columns. Euler's For- 
mula. Empirical Formulae. Rankine's or Gordon's Straight 
Line Formula 108 

CHAPTER IX. 
Cotters: Design of Cottered Joint. — Gib and Cotter. — Taper of the 

Cotter 123 



8 CONTENTS. 

CHAPTER X. 
The Theory of the Connecting Rod '. '. 133 

CHAPTER XI. 
Practical Problem III: Strapped Connecting Rod 138 

CHAPTER XII. 
Practical Problem IV: The Naval Type of Connecting Rod 147 

CHAPTER XIII. 
Shafts and Journals: Torsion. Strength of Shafts. Twist of Shafts. 
Size of Shaft for Given Horse-Power. Strength of Journals and 
Pins. Table of Bearing Pressures. Combined Turning and 
Bending Moments. Couplings and Bolts 172 

CHAPTER XIV. 
Friction and Lubrication of Bearings: General Effect of Friction. 
Dry and Lubricated Surfaces. Nominal or Projected Area of 
Bearing. Work of Friction. Value of /jl. Pressure in Bearing. 
Forced Lubrication. Anti-Friction Metal. Thrust Bearing. . . . 187 

CHAPTER XV. 

Notes on the Design of Crank Shaft, Cross-Head Pins, etc.: Practi- 
cal Problem V 200 

CHAPTER XVI. 
I. H. P. Required. Design of Cylinders, Valve Chests, Pistons, Rods, 
Pipes, etc. — Bureau of Steam Engineering Check for Strenth 
of Steel Pistons 216 

CHAPTER XVII. 
Notes on the Design of Cylindrical Boilers 253 

CHAPTER XVIII. 
Notes on Screw Propellers 291 



ENGINEERING MECHANICS 



CHAPTER I. 

Units, Definitions, Quality of Materials. 

1. The following units will be used tlirougiiout, unless specifically 
stated to the contrary : 

Dimensions follow the usual practice of the Bnrean of Steam 
Engineering, being given in feet and inches, when greater than 2-i 
inches, and in inches up to and including 24''. For obtaining stresses, 
moments, etc., the dimensions are reduced to inches. 

Loads and forces are given in pounds. When the ton is used it is 
always the standard ton of 2240 pounds. 

Stresses are given in pounds per square inch. 

Pressures are in pounds per square inch. 

Velocities are in feet per minute, or feet per second. 

Accelerations are in feet per second per second. 

Work is in foot pounds. 

Power is in foot pounds per minute. 

Speed of rotation is in revolutions per minute, or in angular 
velocity per second. 

Moments (such as bending and twisting or turning moments) are 
in inch-pounds. 

There is frequently a tendency to use the same terms to express 
the forces acting on the parts of a machine or structure and the 
deformations produced by these forces ; that is to sa}^, the distinction 
between stress and strain is not clearly made. In the same way the 
same term is often used to express either a quantity or an intensity. 
This common error may be avoided by constantly having in mind the 
following relations : 

Stress tension pressure shearing stress ^ ,. ., 

^ — r- = — : — , or^ , or ^j . ^ ^ — r- = elasticity. 

Strain extension compressure shearing strain "^ 

Extension (e) = -^^i^Bgi^EiA) or e = } . 
original length (1) 1 

Compressions -C05traction__ 
original length 



10 ENGINEERING MECHANICS. 

2. Strain and Stress. — A strain is any definite alteration of form 
or dimensions experienced by a solid body. It is the change of shape 
or size of a body, especially of a solid, produced by the action of a 
stress; therefore, it is deformation, and it may be temporary or 
permanent. When the strain is temporary and disappears on the 
removal of the force by which it is produced, it is called the elastic 
deformation; when it remains after the removal of the producing 
force, it is a permanent deformation or permanent set. 

Stress is the internal reaction between the molecules of a material 
produced by the application of external loads or forces. Thus stress 
is of the nature of a force, while strain is a cliange of dimension. 

Elasticity is the relation between stress and strain, within the 

elastic limit, and the ratio i'-^''' ^f "'^^^ "^ cross sectional area\ 

[ strain per unit of length J 

the modulus of elasticity. This will be denoted by the symbol E for 
direct tension or compression. Thus : 

E=l. 
e 

Elastic and Plastic Conditions. — An elastic material is one that, 
when deformed or strained by the action of a force, recovers its 
original size and shape when the straining force is removed. A 
plastic material is one which does not completely recover its original 
size and shape on the removal of the straining force, but takes a 
permanent set. In general, every strain consists of (a) an elastic 
deformation, which is proportional to the straining force, and (b) 
a plastic deformation or set. For the materials used in the con- 
struction of machinery and other structures there is a range of 
straining force within which the strain is practically wholly elastic. 
This strain can, therefore, be allowed for in the design, by provid- 
ing that the working stress produced by the forces acting on the 
several parts of the machine does not exceed the elastic limit of the 
material employed. 

Solid materials which possess plasticity exhibit the phenomenon 
of flow under unequal stresses in different directions, much in the 
same way as liquids. This property of flowing is utilized in the 
manufacture of lead pipe, the drawing of wire, the stamping of 
coins, forging, etc. 

Ductility is that property of a material by virtue of which it may 
be drawn out bv tension to a smaller section, as when wire is made 



UNITS, DEFINITIONS, QUALITY OF MATERIALS. 11 



12 ENGINEERING MECHANICS. 

by drawing out metal through a hole in a die plate. During ductile 
extension a material shows a small degree of elasticity together with 
a much greater degree of plasticity. 

Brittleness is the lack of ductility. 

Malleability is that property of a material by virtue of which it 
can be hammered or rolled into the form of a plate. This property 
is very similar to ductility. 

3. The Belation of Stress and Strain. — The relation between stress 
and strain is very clearly shown on the ordinary load-strain or 
stress-strain diagrams, such as are illustrated by Eig. 2. 

Suppose the bar represented by Fig. 1 is subjected to a gradually 
increasing tensile load and at the same time the increase in length 
between the fixed points a and b is measured. 

A diagram in which the ordinates represent stress per unit of 
area differs from one in which the ordinates represent total load, or 
stress per unit of original area of cross section, because, as the load 



X 



./» 



Fig. 1.— Tensile Test Piece. 

is increased, the straining action produces a contraction of area. 
These diagrams are usually made to represent the relation of load, 
or total force, and strain, and Fig. 2 represents such a diagram. 

Let the horizontal axis represent strains and the vertical axis 
represent loads, or stresses ; then the relation between strain and load, 
or stress, will be represented by the curves. Plot extensions to the 
right, compressions to the left ; tensions upward and pressures down- 
ward. The curve for a ductile material, such as mild steel, will then 
be Oabcde for tension, and Ofg for compression. As the loads in 
tension are gradually applied the bar stretches, and up to the load 
represented by the ordinate a^a the curve will be practically a 
straight line. That is, the strains are proportional to the loads, and if 
the load is released the bar will return to its original length. The 
strains are elastic or temporary. At the point a the proportionality 
between load and strain ceases, and the strains increase much more 
rapidly than before, and a part of the strain is now permanent. 
The point a is, therefore, the limit of proportionality or the true 
clastic limit. When the material being tested is subjected to rolling, 
or forging in the process of manufacture, there is generally a second, 



UNITS, DEFINITIONS, QUALITY OF MATERIALS. 



13 



well-defined point b, at which the test piece takes a great increase of 
strain from b^ to c^^ without additional stress. This marks the yield 
point, and is commercially referred to as the elastic limit. As the 
load is further increased the strains are rapidly increased, as from 
c to d, until at d the maximum load is reached ; further load beyond 
this point causes a local reduction of area of the test piece, so that it 
can no longer sustain the maximum load, and it finally breaks with 









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Fig. 2.— Typical Stress-Strain Diagram. 



a load ee^ less than the maximum dd. The stress due to the load dd. 
per unit of area of the original section of the test piece is called the 
ultimate strength, or hreaJcing stress of the material. 

The intensity of the stress per unit of area of the reduced section 
of the test piece increases constantly up to the point of rupture, and 
is much greater than the so-called breaking stress. Under compres- 
sion the curve is of the nature of Ofg, having a much less defined 
yield point at f than under tension. Under coinpression the cross- 
sectional area of the test piece increases, and, consequently, the 



14 ENGIJ^EERII^G MECHANICS. 

straining action increases up to the point of final breaking down of 
the material. 

For snch materials as cast iron and the hard varieties of steel the 
load-strain curve has the general form shown on the diagram, there 
being no yield point. Snch material breaks with much less stress 
than those having well-defined yield points, and is brittle. Brass 
and bronze usually do not exhibit an elastic limit, or yield point, 
but are subject to great plastic deformation before final rupture, 
and are tough. 

4. Tlie Forces Acting on the Parts of a Machine. — All materials 
used in construction are more or less elastic, and, therefore, any part 
of a machine under the action of a load or force must change its 
form. This change of form may be very small and temporary; or 
it may be large and permanent; and if the load is sufficiently great 
rupture will occur. The character of the straining action and the 
stresses produced depend on the direction and point of application 
of the load, and on the form and method of support of the loaded 
part. Thus the load may produce tension, compression, shearing, 
bending, torsion, or a combination of these actions. While tension 
and compression cannot exist at one time between any pair of mole- 
cules, yet, in the case of bending, tension and compression do occur 
in different sets of fibers of the same part. This action will be more 
fully explained later. In tension, compression and bending the 
molecular stresses act normally to the planes separating the inter- 
acting molecules: that is, the stresses increase or decrease tbe 
distances between these molecules along the lines joining them. In 
shearing and torsion the displacement of the molecules is tan- 
gentially to the planes between the molecules. In pure shear the 
motion is rectilinear, while in torsion the adjacent molecules have a 
relative motion about an axis. Theoretically, only two kinds of 
strain exist: viz, elongation (compression being considered as nega- 
tive elongation) and shearing. Similarly there are only two kinds 
of stresses : ^dz, direct or normal and tangential or shearing. But 
for practical convenience it is desirable to treat torsion and bending 
separately as elementary stresses. 

The forces acting on the parts of a machine are summarized by 
TJnwin as follows: (1) The useful load due to the effort trans- 
mitted from the driving to the working point to accomplish useful 
work. (2) Prejudicial resistances due to friction in the machine 
itself or work expended otherwise than at the working point. (3) 



UNITS, DEFINITIONS, QUALITY OF MATERIALS. 15 

The weight of parts of the machine. (4) Eeactions of inertia due to 
changes of velocity of parts of the machine. (5) Centrifugal forces^ 
due to changes of direction of motion of parts of the machine. (6) 
Occasionally there are stresses due to constraint, preventing expan- 
sion with changes of temperature. 

The total action, on any member of a machine^ of these forces 
may be called the total straining action on that member. The 
relative importance of the various straining actions is very different 
in different cases, and generally some of them are small enough to be 
neglected. The problems of designing are then simplified. 

In each part of a machine the straining action varies with the 
fluctuations of tlie useful load and with the variations of position 
and velocity of the different parts of the machine. Each member 
must be capable of resisting the maximum straining action on that 
part of the machine. For each part of the machine, therefore, it is 
necessary to consider under what conditions the straining action is 
greatest. If in consequence of changes of position or velocity the 
straining action produces stresses of different kinds at different 
times, the member must be capable of sustaining the maximum 
stress of each kind. Lastly, as will be more fully explained pres- 
ently, the amount of variation of the straining action affects the 
endurance of the material^ and, therefore, requires also to be con- 
sidered. 

Questions and Problems. 

Distinguish between stress, strain, tension, extension, pressure, 
compression. Define elasticity and the modulus of elasticity of 
extension and compression. Discuss the following properties of 
materials : elastic and plastic conditions, ductility, brittleness, 
malleability. 

Show, by means of diagrams, the relation between stress and 
strain for ductile and brittle materials, explaining clearly the fol- 
lowing: true elastic limit^ yield point, ultimate strength. 

Discuss the forces acting on the parts of a machine. 

Problems. 

1. A steel rod, 3 J feet long, 24- square inches sectional area, 
reaches the elastic limit at 125,000 pounds, with an elongation of 
0.065 inches. Find the stress and strain at the elastic limit, and 
the modulus of elasticity and express each in its proper units. 



16 ENGINEERING MECHANICS. 

2. A piston rod is 10 feet long and 7 inches diameter. The 
diameter of the cylinder is 70 inches, and the effective steam pressure 
100 pounds per square inch. Find the stress produced and the total 
alteration in the length of the rod for a complete revolution, 
E = 30,000,000 inch-pound units. 

3. A brass pump rod is 5 feet long and 4 inches diameter, and 
lifts a bucket 28 inches diameter on which is a pressure of 6 pounds 
per square inch, in addition to the atmosphere. The pressure below 
the bucket is 2 pounds absolute. What is the stress in the rod and 
the total extension per stroke? E = 9,000,000 inch-pound units. 

4. A round bar of mild steel, 18 feet long and 1-J inches diameter, 
lengthens 1/16 inch under a pull of 7 tons. Find the intensity of 
the stress, and the value of E. , 

5. A hollow cylindrical cast-iron column is 10 inches outside and 
8 inches inside diameter, and 10 feet long. How much will it 
shorten under a load of 60 tons? Assume E = 18,000,000 inch- 
pound units. 



CHAPTER II. 

The General Peoblems of Design. Straining Actions. 
Factors of Safety. 

5.* The General Prohlem of Designing Machinery may be divided 
into two parts : 

(1) Design of the mechanism to give the required motion. 

(2) Proportioning of the parts so that they will carry the neces- 
sary loads due to transmitting the energy, without undue distortion 
or practical departure from the required constrained motion (1) is 
the application of the principles of mechanism. Very frequently 
some well-known and thoroughly tried mechanism can be adopted 
for the problem under solution. In other cases, where some new 
tj^pe of machine is to be built, the problem becomes practically one 
of invention. 

The proportioning of the various parts may conveniently be 
divided into two parts : 

(a) Solution as a whole of the energy and force problem of the 
mechanism. 

When the type and proportions of the mechanism have been 
fixed the relative velocity of any point in the mechanism may be 
found. If, then, the energy which the mechanism must transmit is 
known, it is possible, in general, to find the forces acting, from the 
law of conservation of energy; or the product of velocity multiplied 
by force is constant throughout the train. 

(b) Assigning the dimensions of the various parts based on the 
forces acting on them. 

When the forces acting can be determined it would seem easy 
to choose the material and assign proportions for a machine part, 
based on the laws of mechanics or strength of material, and such 
is the case when the stresses are simple and the conditions fully 
known. Thus a machine member, subject to simple tension within 
known limits, can be intelligently proportioned in this manner. But 
in many cases the forces acting are very complex, the theoretical 
design is not always clear, and our knowledge of materials and their 

* From Kimball and Barr. 



18 ENGINEERING MECHANICS. 

laws is limited in many respects. Eecourse mnst, therefore, often 
be made to judgment or to empirical data, the result of experience. 
Even when the conditions are clear, theoretical design must always 
be tempered with practical modification and by constructive con- 
siderations, etc. The logical method of proportioning machine ele- 
ments where theory is applicable is, therefore, as follows : 

(1st) Make as close an analysis as possible of all forces acting and 
proportion parts according to theoretical principles. 

(2d) Modify such design by judgment and a consideration of the 
practical production of the part. 

In the case of details and unimportant parts, judgment and 
empirical data are commonly the best guides. 

Summing up, then, the logical steps in the design of a machine 
are as follows : 

(I) Selection of the mechanism. 
(II) Solution of the energy and force problem. 

(III) Design of the various macliine members so they will not 
unduly distort or break under the loads carried. 

(IV) Specification and drawing. 

The last step, specification and drawing, is a necessary and im- 
portant adjunct to the process of design ; it is a powerful aid to the 
designer's mental process and is the best way of showing the work- 
man what is to be done to construct the machine in question and 
also of making a record of what has actually been done. It is not 
machine design of itself, however, as machines may be designed and 
built without any drawings. It is, nevertheless, an indispensable 
part of the designer's equipment. Very often written specifications 
accompanying the drawings are not only useful, but necessary. In 
fact, the highest skill on the part of the designer is often needed to 
clearly and fully specify in writing just what is to be done, as the 
writing of specifications presupposes the most intimate knowledge 
of theory of design, and selection of materials. 

From the above it is seen that the courses already completed, in 
princij)les of mechanism, mechanical processes, strength of mate- 
rials, etc., are a necessary ground work before advancing to machine 
design. 

6. The Effect of Temperature. — Certain parts of machines or 
other structures must work under high temperatures, as, for in- 
stance, boilers, steam valves, etc., so the effect of such temperatures 



THE GENERAL PEOBLEMS OF DESIGN. 



19 



must be taken into consideration. Fortunately, iron and steel do 
not suffer very appreciable change of strength under the tempera- 
tures ordinarily reached in practice. The following tables show the 
relative tenacity of wrouglit iron and steel at different temperatures, 
compared to that at ordinary temperature^ the tenacity at ordinary 
temperature being taken as 100 (from Unwin). 



TABLE 1.— WROUGHT IRON. 





213° 




600<: 


> i 7500 






lonno 


1200° 


RpilfltivA tfinaf»if.v 


104 


116 I 96 

1 


76 ; 

1 


40 








TABLE 2.— MILD STEEL. 






Temp. F 


- 4° 
106 


213° 
103 


400° 


570° 1 
123 


750° 


900° 
49 


1100° 


Relative tenacity. . . 




133 


86 




28 



Professor Morley summarizes the results of experiments on the 
effect of temperature as follows : 

"(1) The tenacity (a) at ordinary temperature falls off with 
increased temperatures until between 200° and 300° F., when it is 
something of the order of 5 per cent less than at 60° F. (b) It 
rises from this temperature to a maximum value at some tempera- 
ture between 400° and 600° F., when it is something of the order 
of 15 per cent more than 60° F. (c) It falls continuously with 
further increase of temperature. 

"(2) Tlie elastic limit falls continuously with increase of tem- 
perature. 

"(3) Tlie elongation (a) falls with increase of temperature above 
the normal to a minimum value in the neighborhood of 300° F., 
and then (b) rises again continuously with increase of temperature. 



"(4) The modulus of direct elasticity (E) decreases steadily 
with increase of temperature, metals which give a value of about 
13,000 tons per square inch at atmospheric temperature falling to 
about 12,000 tons per square inch at 500° F. 

'' Low Temperatures. — Experiments on a very mild steel at very 
low temperature show progressive increase of tenacity with decrease 
of temperature ; while the elongation practically vanishes, the mate- 



20 ENGINEERING MECHANICS. 

rial behaving like a very brittle substance. On return to ordinary 
temperatures no permanent change from the original properties 
is observed." 

The loss of strength with increase of temperature of copper and 
the copper alloys is much greater than that of iron and steel. ITnwin 
gives the following for such materials: 

f = a-b(t-60)^ 

where f is the tenacity at t° F., in tons per square inch, and a and 
b are constants having the following values : 

TABLE 3. 

a b 

Copper 14.8 .000014 

Rolled yellow brass 24.1 .000028 

Rolled delta metal 31.3 .000041 

Rolled Muntz metal 14.7 .000029 

Cast gun metal 12.5 .000021 

Cast brass 12.5 .000024 

Cast phosphor bronze 16.1 .000026 

7. Stress due to Change of Temperature. — l^early all metals ex- 
pand when heated and contract again when cooled. The expansion 
of the metal per unit of length for a rise of temperature of 1°, is 
the coefficient of expansion, which will be denoted by a. 

Then the following are the values of a per degree Farenheit : 

TABLE 4. 

Hard steel a = .0000074 

Soft steel a = .0000065 

Cast iron a = .0000062 

Wrought iron a = .0000068 

Copper and its alloys a = .0000010 

Suppose a bar of metal to be heated and then to be securely fixed 
at the ends, so as to entirely prevent its contraction when cooled, 
there will be a strong pull on the supports^ producing a tensional 
stress in the bar. Let the original length of the bar be 1, and the 
rise of temperature be t° F., and a be the coefficient of expansion; 
then the length when hot will be 1(1 + at) . After the bar has cooled 
to the original temperature (but its contraction having been wholly 
prevented) there is a strain per unit of length of at. Let f be the 
stress produced per unit of area of cross section. Then, remember- 



THE GENERAL PROBLEMS OF DESIGN. 21 

ing the definitions of elasticity and the modulus of elasticity, we 
have 

E='^^^=i-.-.f = atE. 
strain at 

8. The Nature of Loads. — Steady or Dead Load. — A steady load 
is one which is invariable both in direction and amount during the 
life time of the machine, and which, therefore, produces a per- 
manent straining action. The weight of the part of a machine is 
an example of a steady load. 

Variable or Live Load. — A variable load is one which is alter- 
nately put on and taken off^ and which produces a straining action 
constantly varying in amount. The action of steam on the piston 
of an engine, and the passage of a train over a bridge are examples 
of such a load. 

The effect of a steady load is generally easily calculated and the 
dimensions of the part to resist its action can be very exactly deter- 
mined, and, consequently, a low factor of safety may be safely 
employed. The live load is more difficult to allow for, and, more- 
over, its effect is more destructive on the parts resisting it, than the 
steady load. A higher factor of safety must, therefore, be used in 
determining the dimensions of the parts subject to such load. 

Sudden and Impulsive Loads. — A sudden load is one applied 
without velocity, but at one instant, and continues to act during the 
deformation produced. Just at first the load is not balanced by the 
stress, so that it acquires a certain amount of kinetic energy, while 
the resisting part is being deformed, and this energy is effective in 
increasing the deformation beyond that due to the same steady 
load. If the greatest stress thus produced is within the elastic limit 
of the material, the amount of this stress, momentarily, is double 
that due to the same load applied gradually, or resting on the part. 

If a heavy body impinges with velocity and thus possesses kinetic 
energy the stress produced is much greater than that due to the 
same body resting on the part. 

Such a load is said to be an impulsive load, and the stresses are 
said to be due to shock. In order to find the value of the stress in 
such a case, the work of deformation is equated to the kinetic 
energy of the impinging load. 

This subject will be further treated later (see Art. 13). 



22 ENGINEERING MECHANICS. 

9. Fatigue. — It is a well-known fact that parts of machines have 
broken, after greater or less length of time, under loads which have 
previously produced no bad effects. This is especially the case when 
the load has been such as to produce alternations of tensile and com- 
pressive stresses. Thus piston rods and crank shafts have suddenly 
broken under usual working conditions. The only reason that can 
be assigned for such failures is that the material under such alterna- 
tions of stress suffers a deterioration, which renders it incapable of 
resisting a stress previously safe. This deterioration is called 
fatigue. Microscopic examinations of specimens under test have 
led to a theory to account for the effect of fatigue. It was found 
that the crystalline particles of the material are changed in shape 
by slips at the plane of cleavage of the cry'stals. These slips grad- 
ually increase in number, and becoming massed together produce 
cracks. When the specimen is under reversed stresses these cracks 
are much more rapidly developed by the loss of cohesion due to the 
grinding action on the cleavage planes, along which slipping to and 
fro takes place. 

Another fact having a bearing on this subject is the well-known 
variation of elastic limit of ductile materials. Thus if a specimen 
is tested under tension, and the test is discontinued before rupture, 
and the test then repeated, it is found that the elastic limit has a 
higher value at the second test — but the ductilit}^ is reduced, and 
the material is more brittle than before. In the same way, if a 
specimen is subject to a compressive test, and then a second test is 
applied to the specimen, but this time, under a tensile load, the 
elastic limit is found to be much lowered. 

10. Experiments on the Action of Live Loads. — ISTumerous ex- 
periments have been made to determine the effect of live loads, and 
of loads causing a reversal of stress. Of these Wohler's researches 
are generally accepted as giving the most satisfactory working 
hypothesis. His experiments consisted of a series of a large num- 
ber of tests in which the stresses were alternately produced and 
wholly or partially removed; and others in which the stresses were 
alternately tension and compression. Also series of tests under 
torsional stresses. The stresses applied at first were very large, in 
fact, nearly up to the limit of the statical breaking strength, and 
such stresses produced fracture after a small number of repetitions 
of load, but the number of repetitions before fracture occurred 
rapidly increased as the limit of the stresses was reduced, and. 



THE GENERAL PROBLEMS OF DESIGN. 



23 



finally, for a very low limit of stresses, the number of repetitions 
necessary to produce fracture approached an infinite number. 

These experiments show that the stress, at which a material frac- 
tureS;, when subject to varying strains, depends on the range of 
variation of stress and on the number of repetitions of change of 
load. That is to say, the safe working stress is lower the greater 
the range of stress produced and the greater the number of repeti- 
tions of loading. For example, the crank shaft and the piston rod 
of an engine suffer a complete cycle of reversal of stress for each 
revolution of the engine. Therefore, such parts, in order to run 
safely for indefinitely long periods, must be designed for much 
smaller working stresses than would be allowable for the same mate- 
rial under a steady load. 

The following table shows very clearly the general nature of the 
results obtained by Wohler: 

TABLE 5.— RUPTURE OF WROUGHT IRON BARS BY TENSION. 

Rupture by 1 application of 55,000 lbs. per sq. inch. 

800 applications of 51,500 " 

" 107,000 " " 47,000 " " 

" 341,000 " " 42,000 " " 

" 481,000 " " 38,000 " " " 



A Sample of Spking Steel, Subjected to Bending, Broke: 

Wlien subjected to 81,000 applications of 95,000 lbs. per sq. in 

" 154,000 " " 85,000 

" 210,000 " " 75,000 

" 472,000 " " 65,000 

" 539,000 " " 58,000 



1,165,000 



53,000 



11. Algebraic Equation for Woliler's Law. — Let C be the statical * 
breaking strength of a material, and suppose that the stress in a 
bar, or a part of a machine, varies from a value represented by iniax 
to a value of imin, so that the range of stress is E^f^a^ — f,„in. To 



* Some writers, particularly those of Germany, give the breaking 
strength different names, according to the conditions under which the 
piece is placed. Thus the stress at which rupture occurs under a steady 
or very gradually applied load is called the statical breaking strength. 
When the bar returns to its original condition after each application 
of a repeated load, and if the stresses are all of the same sign (that 



24 ENGINEERING MECHANICS. 

use this expression call tension + and pressures — . Then if the 
two stresses are of different sign, the range, R= [fwa^— ( — fmin)] 
= imax + imin- If the number of changes of load is indefinitely 
great, fracture will occur for a value of tmax less than C, and so much 
smaller the greater the value of E. In other words, the breaking 
strength is less than C. Therefore, in designing for a var3dng load 
the working stress mu.st be taken at a value of f less than C, and 
this value depends upon E. imax is the breaking strength of a 
material under a variable load ranging between the limits of imaw 
and ±imin repeated an indefinite^ great number of times, imin is 
+ if the stress is the same kind as imax and — if of an opposite 
kind, and, further, imin is supposed to be less than imax- Then 
the range ^ — imax^i-min, the upper sign being taken if the stresses 
are of the same kind and the lower if they are opposite. Thus E 
will always be -\- . 

The results of Wohler's experiments can then be expressed by 
the equation 

f«a.-|+VC^-kEC. (1) 

If E = 0, then imax = 0, the load being steady, k is a constant, 
such that when given suitable values, the decrease of imax, as the 
value of E increases, is made to conform with the values observed 
in Wohler's experiments. For the more ductile varieties of iron 
and steel the average value of k is 1.5, and for the harder varieties 
k = 2.0. 

There are three principal cases to consider: 

(a) The load is steady; then E = 0. 

(b) The load is entirely removed and replaced; then f^i„ = and 

(c) The load produces alternately a tensile and a compressive 
stress of equal magnitudes; then imax and imin are equal and of 
opposite sign, and E — 2f,na«. 

is, all tensions, compressions, or shearing in one direction), the greatest 
stress that can be sustained for a given number of repetitions is called 
the primitive strength. If the stresses are alternately of opposite signs, 
that is alternately tension and compression, or shearing in opposite 
directions, the maximum stress which can be sustained is the vihration 
strength. 



THE GENERAL PEOBLEMS OF DESIGN. lr> 



Fornmla (1) then takes the following forms for these eases: 

Greatest Least Range of Value of f> 

stress, stress. stress, '"" 

(a) fn^a. tmax R = f,.a.r = ^ + V C^ - k X X C = C ; 



(b) i-max ^ -tt — ^max ^maa — — jy -\- \' kj Ki-maxy^ 'y 



2 

n 

"2 



f^-|^J=C^-kf,„«,.C, 



which reduces to 

= 2C(VF+l-k), 
2L 



(C) tmax ^max ^ — ^^max imax'= ,-, -\- v Kj" — /^KUI: 

f — 



Q I V ^ "^'■^^maxf 

2k* 



By giving k a value of 1.5, which is its average value for the kinds 
of iron and steel most commonly nsed for construction, the above 
equations reduce to the following : 

( a j tmax =^ ^ j 

(b) f^a. = 0.60540; 

That is to say, in round numbers^ the stresses which a bar of given 
material can sustain are in the ratio of 3:2:1 for the three kinds 
of loading. 

12. Factor of Safety. — It is now evident that the parts of a 
machine must be so designed that the stresses produced in each are 
within the elastic limits of the materials. The worMng stress is the 
stress at which any machine part is designed to operate^ and the 
ratio of the hreaking stress to the ivorking stress is the factor of 
safety. It is not suflQcient generally that the working stress be kept 
only below the elastic limit, but the size of the part must be such 
that the strain produced is so small as not to throw it out of align- 
ment, which might produce unnecessary friction, or even inter- 
ference with other parts. A particular part might easily be made 
amply strong to resist the forces acting on it, but yet it might dis- 
tort to such an extent as to render it unfit for its purpose. Thus 
it is seen that considerations both of strength and of stiffness must 
be employed in determining the sizes of the various parts of 
machinery. 



26 EN'GINEERING MECHANICS. 

A distinction must be made between the apparent factor of safety, 
as defined above, and the real factor of safety. The latter is the 
ratio of the carrying strength, or the maximum allowable stress as 
determined by Wohler^s law, to the working stress. Thus, suppose 
we use a factor of safety of 3, to insure that the stress will be wdthin 
the elastic limit, and to cover defects of material, and uncalculated 
effects, then the apparent factors of safety will be approximately : 

For a steady or dead load, factor of safety = 3. 

For a load alternately removed and replaced, factor of safetv 
= 3x2 = 6. 

For a load producing stresses alternately in opposite directions, 
factor of safety = 3x3 = 9. 

The actual selection of the proper factor of safety requires the 
exercise of the highest skill and trained judgment on the part of 
the engineer. The following from Kent, on " Factors of Safety/' 
shows the general nature of the considerations necessary : 

" The selection of the proper factor of safety or the proper maxi- 
mum unit stress for any given case is a matter to be largely deter- 
mined by the judgment of the engineer and by experience. No 
definite rules can be given. In general, the following circumstances 
are to be taken into account in the selection of a factor : 

'^ 1. When the ultimate strength of the material is known within 
narrow limits, as in the case of structural steel when tests of samples 
have been made, when the load is entirely a steady one of a known 
amount, and there is no reason to fear the deterioration of the metal 
by corrosion, the lowest factor that should be adopted is 3. 

" 2. When the circumstances are modified by a portion of the load 
being variable, as in floors of warehouses, the factor should not be 
less than 4. 

" 3. When the whole load, or nearly the whole, is apt to be alter- 
nately put on and taken off, as in suspension rods of floors of 
bridges, the factor should be 5 or 6. 

"4. When the stresses are reversed in direction from tension to 
compression, as in some bridge diagonals and parts of machines, the 
factor should not be less than 6. 

" 5. When the piece is subjected to repeated shocks, the factor 
should be not less than 10. 

" 6. When the piece is subject to deterioration from corrosion, 
the section should be sufficiently increased to allow for a definite 



THE GENERAL PROBLEMS OF DESIGN. 



27 



amount of corrosion before the piece be so far weakened by it as to 
require removal. 

" 7. When the strength of the material, or the amount of the 
load, or both, are uncertain, the factor should be increased by an 
allowance sufficient to cover the amount of the uncertaint)^ 

" 8. When the strains are of a complex character and of uncer- 
tain amount, such as those in the crank-shaft of a reversing engine, 
a very high factor is necessary, possibly, even as high as 40, the 
figure given by Eankine for shafts in mill work.^^ 

In the absence of experience the following table of factors of 
safety may be used : 

TABLE 6.— FACTORS OF SAFETY. 



Material. 


Dead 
load. 


Repeated stresses in 
one direction. 


Reversed 


stresses. 


Gradually 
applied. 


Suddenly 
applied. 


Gradually 
applied. 


Suddenly 
applied. 


Wroug:ht iron, steel and 
ductile metals 


3 

4 


5 

6 


10 
12 


6 
10 


13 


Cast iron and brittle 
metals 


20 



Questions and Problems. 

Discuss the general problem of designing a machine. 
Explain the effect of temperature on materials and the stresses 
due to change of temperature. 

Define steady load, live load, sudden load, impulsive load, fatigue. 
Explain Wohler's experiments on the action of live loads. 

R 



Given Wohler's formula, f,. 



-f-VC^-kRC, explain the 



meaning of the symbols, and show how the formula becomes modi- 
fied (1) for a steady load, (2) for a repeated load, (3) for a vibrat- 
ing load. Assuming k = 1.5, determine the value of fmax in terms 
of C for each of the above cases. 

Define factor of safety; distinguish between real and apparent 
factors of safety. Discuss the general considerations leading to the 
selection of a factor of safety. 



Problems. 

1. A driven rivet cools through 650° F. Coefficient of expansion 
.00001, E = 25,000,000. Find the maximum possible tensile stress 



28 ElI^GINEERING MECHANICS. 

per square inch. In practice would the actual stress be as great as 
this? Why? 

2. A bar of steel 1 inch diameter and 10 feet long, is heated 
100° F, and the ends are then firmly gripped. Find the tensional 
stress per square inch, and the total pull of the bar, when it returns 
to its original temperature, if the end supports of the grips are 
pulled 1/40 inch nearer together. Coefficient of expansion for steel 
.0000062; E = 26,000,000 lbs per square inch. 

3. Two walls, 25 feet apart, are stayed together by a steel bar 
1 inch diameter. The bar passes through plates, and is set up by 
nuts at each end. If the nuts are screwed up to the plates while the 
bar is at a temperature of 300° F., find the total pull exerted when 
the bar cools to 60 F. (1) if the ends do not yield, (2) if the total 
yielding at the two ends is i inch. Coefficient of expansion for 
steel = .0000062, E = 27,000,000 pounds per square inch. 

4. In the test of a certain sample of steel it is found that the 
stress at the elastic limit is 71,000 pounds per square inch, and 
that the ultimate strength is 118,000 pounds per square inch. What 
factor of safety must be used in order to bring the working stress 
within the elastic limit ? 

5. Plot a curve showing the effect of temperature on the tenacity 
of copper. (Suggestion, let the vertical ordinates represent f, the 
tenacity and the abcissse represent temperatures.) Art. 6. 

6. Plot a curve showing the effect of temperature on phosphor 
bronze (Art. 6). 



CHAPTER III. 



Resistance of Materials to Straining Actions. 

resilience, sudden and impulsive loads^ tables of strength, 
etc. tension, compression and shearing. 



^ 



13. Resilience. — ^This term is commonl}: 
of a strained bod}^ in springing 
back when the straining force is 
removed. As used technically, in 
mechanics, it means the work done 
in straining a material np to the 
elastic limit. Fig. 3 represents the 
portion of a stress-strain diagram, ^ ^ 
within the elastic limit. If a bar ^ 
of metal is subjected to a gradually K 
increasing stress, the strain increases 
in proportion to the increase of 
stress, within the elastic limit. The 
work done in producing the strain 
is represented b}^ the shaded area 
OAB of Fig. 3. 



used to express the action 




Fig. 3. 



Let A=the stretch. 

f = the stress per unit of cross-sectional area. 
l=:the original length of the bar. 
A = Area of section of the bar. 
E = Modulns of elasticity. 



Then y =6 



E: 



strain per unit of length, and 
stress f ]f 



strain 






The work done in straining the bar of sectional area A is, then, 



from Fig. 3. 



30 



ENGINEERING MECHANICS. 



Work per unit of area = area of triangle OAB=^fA. 



Total work: 



Af\ 
2 



f^Al 
2E 



Al f^ 
2 ^ E 



Al P 

Now —^ is ^ the volume of the bar, and ^^r is called the modulus 

of elastic resilience: therefore, the work done = |- volume of bar 
X modulus of elastic resilience. 



,h^H 



14. The Effect of Sudden and Impulsive Loads. — Suppose that a 
load W i& gradually applied to a bar of sectional area A; and 

that the bar stretches an amount A 

(from to B, Fig. 4). The stress 

■produced will be zero at 0, as it is 

applied gradually, increasing to f at 

W 

— r- ; and the mean stress 
A 

W 




B, and f: 



during the operation is fmean = 



2A 



Now suppose the load W is sud- 
denly applied and that the bar 
stretches from to E (Fig. 4). 

The work strain caused by weight 
W falling through distance \=Wk 
n=fAA. 



When the weight reaches B only a part of the energy stored in the 
weight has been expended in stretching the bar, and the remainder 
is still stored in the weight (as kinetic energy) and this energy is 
absorbed in further extending the bar from B to E. The work 

absorbed by the bar in stretching an amount A is — ^ . The energ}^ 

remaining in the weight when it reaches B is, then, the difference 
between the total energy and the energy absorbed by the bar. This 

difference is f A\ — -^y- = —^ . 

The bar will continue to stretch until the energy absorbed by it is 
equal to the total energy in the weight. 

This will occur when the OEC = the area OFDE, and then the 
area ADC = the area OB A. 



EESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



31 



When the weight reaches E the strain 
and, therefore, also the stress is double that 

at B. Thus a suddenly applied load pro- ///^///////yy////y///y/, 
duces twice the stress, and twice the strain 
produced by the same load gradually ap- 
plied. 

In Fig. 5 suppose a weight W falls 
through a height h and brings up on a stop 
in such a manner as to produce an axial 
tension in a bar. As before 

Let 1 = original length of bai. 
A = the stretch. 
f = the stress produced. 

Then the total work done by the weight 
= W(h + A) : The work absorbed by the bar 
Al ^2 






Fig. 



is ^f AA = V X ^ • ^ <^^^'' since all the connections are supposed 
2 hi 

to be absolutely rigid, the two expressions for work must be equal, or 



W(h + A)=^^~X-j. 



2E 



X volume of the bar 



or f^: 



2EW(h + A) 
volume of bar ' 

from which f can be easily calculated when E for the given material 
is known.* 

If, in the above equation, we put h = 0, we have the case of the 
sndden load. Then 

2EwA 2 EWX _9 W; g Jy 
volume ~ Al ^ A 1 



p 



2^Ee = 2^f 
A A 



then t = 2 



W 



or the stress for a sudden load is double that produced by the same 
load applied gradually. 

15. Table of Strength and Coefficients of Elasticity. — The follow- 
ing table, Table T, taken from Kimball and Barr, with a few altera- 



* When h is large compared to X, which is usually the case, \ may be 
dropped, and the equation becomes 

2EWh 



f2 = 



volume 



32 



ENGINEERING MECHANICS. 



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RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



33 



tions, shows the ultimate and elastic strength of different materials 
commonly used in engineering, when the stress is simple tension, 
compression or shearing. The direct coefficient of elasticity has 
been defined previously. The coefficient of transverse elasticity, 
frequently called the coefficient of rigidity, is the ratio of the shear- 
ing stress is per unit of area, to the distortion n, the distortion being 
measured by the tangent of the difference of the angles of an orig- 
inally square particle before and after the stress is applied. 

In using such a table as the following, it must be understood that 
the values given are general averages and an amount of judgment 
must be exercised in deciding how far such averages may be adopted 
for any particular case. 

Note. — In the solution of problems the values given in Tables 
7 and 8 may be used, when the working stress is not given in the 
problem. 



TABLE 8.— WORKING STRESS. 
A. — Steady ok Permajntent Load. 



Material. 



Cast iron 

Bar iron 

Plate iron, with grain... 
Plate iron, across grain. 

Steel, mild 

Steel castings 



13,000 to 
17,000 
8,000 to 
13,000 
Bronze, phosphor | 10,000 



Kind of stress. 



Tension, 
ft 



4,300 
15,000 
13,500 
13,000 



Gun metal 

Rolled copper. 
Brass 



4,200 
0,000 
3,000 



Compression. 

fc 


Bending. 

fb 


Shear. 

fs 


13,000 
15,000 


6,000 to 
8,000 
15,000 


4.000 
13,000 

10,000 


13,000 to 
17,000 

13,000 to 
16,000 


13,000 to 
17.000 

10.000 to 
14,000 


10,000 to 
13,000 

7,000 to 
13,000 

7,000 

2,400 



Torsion. 



4,000 to 
6.000 
7,500 



8,000 to 
13.000 

7.000 to 
13,000 

4.200 



34 



ENGINEERING MECHANICS. 



TABLE 8 (Continued). 
B. — Load Varying Frequently from o to a Greatest Value. 





Kind of stress. 


Materia], 


Tension, 
ft 


Compression. 

fc 


Bending, 
fb 


Shear. 

fs 


Torsion. 




2,800 

10,000 

9,000 

8,000 

8,000 to 

12,0U0 

6,000 to 

8,000 

6,500 

2,800 

3,000 

2,000 


8,500 
10,000 

8,500 to 
12,000 

8,000 to 
10,500 


4,000 to 
6,500 
10,000 

8,500 to 

12,000 

6,500 to 

9,500 


2,800 
8,000 

6,500 

6,500 to 
8,500 

4.5U0 to 
8,000 

4,500 

.... 
1,600 


2 600 to 


Bar iron 

Plate iron, with grain 

Plate iron, across grain... 

Steel, mild 

Steel castings 


4,000 
5.000 

5,500 to 
8.000 

4,500 to 
8,000 

2,800 












Brass .... 









TABLE 8 (Continued). 
C. — Load Producing Alternate Stresses of Opposite Sign. 



Material. 



Cast iron . . . 

Bar iron 

Steel, mild 

Steel castings 
Gun metal — 



Tension and 
compression. 



1,400 
5,000 

4,500 to 
6,000 

2,500 to 
4,000 

1,400 



Bending. 1 Shear. 



2.000 to 
2.500 
6.000 

4,500 to 
6,000 

3,500 to 
5000 



1,400 
4,000 

3,500 to 
4.500 

2.500 to 
4,000 



Torsion. 



1,200 to 
2,000 
2,500 

2,500 to 
4,000 

2,500 to 
4,000 



RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



35 



16. Resistance to Simple Tension. — A bar is in simple tension 
when the load tends to elongate the bar, and the stress acts along 
lines parallel to the axis. 

Fig. 6 represents a portion of a 
bar on which a longitudinal pull 
is acting. The resultant of the 
load is supposed to act along the 
axis of the bar, and the load will 
then produce a normal tensional 
stress, uniformly distributed over 
a cross section AB. The stress on 
an}' oblique section CD will also 
be uniformly distributed and will 
consist of a direct stress of tension 
normal to the inclined section and 
a shearing stress tangential to the 
section. Let F be the load in 
pounds, and A be the area of the 
normal section in square inches. 
Then the stress on AB will be a 
tension of intensity 
F 



tt^ 



lbs. per sq. in. (1) 



Let the section CD be inclined 
at the angle to the normal sec- 
tion. Then the stresses on CD 
can be found as follows : 

The area of the inclined section 
is A sec 6 — and the intensity of 
the stress, in its original direction 
(along the axis of the bar), over 
the inclined section is 




—- cos = it cos 



A sec ^ ~ A 
Then the normal stress 

fn = fi cos X cos (9 = ff cos -6 
and the tangential stress 

f, — ft cos xsin <9 



(3.) 
(3) 



36 



ENGINEERING MECHANICS. 



Thus it is evident that the stress on any oblique section is less 
than on a normal section^ also the shearing stress is greatest on a 
section inclined at an angle of 45° to the normal, and is, then, equal 
to half the tension on a normal section.* Therefore, the normal 
stresses only need be taken into consideration unless the resistance 
of the material to shearing is less than one-half its resistance to 
tension. 

In practice, the problem is generally to find the cross section 
necessary to carry the given load F. A value for the working stress 
it in tension is selected, from Table 8, depending on the kind of 
material to be used and the nature of the load. Then the normal 
cross section of the part is 

A=f. (5) 

The effect of a tensional load on a bar of 

length 1 and diameter d, is to produce an 

elongation A and a contraction 8. Within the 

elastic limit the extension per unit of length 

X J • TP stress U 

IS e = — and since E = - — ^ = — 

1 strain e 

have 

e=A_ 11. 
1 ~ E 

In the same way the contraction per unit 

of width is 



'r-d- 



we 



(6) 



17 



•^^^d 



(7) 



Fig. 7 represents the bar, the full lines 
representing its unloaded condition, and the 
broken lines, after the load is applied. 

17. Poisso7i's Ratio. — It is found by ex- 
periment that the lateral contraction (or ex- 
pansion, when the load is one of compression) 
is proportional to the change in length of the bar ; that is to say, the 
ratio between the two is a constant, for a given material. This 

constant is usually denoted by — and is called Poisson's ratio. 
J -^ m 



Fig. 7. 



* This is obtained by differentiating equation (4) with respect to 0, 
placing the first derivative = 0, and solving for 6. 



RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



37 



(8) 



Thus 

A _ ^ - ^ 

m "~ X/1 ~ e 

Poisson's ratio varies for different materials from ^ to -^ and its 
average value for metals is about j-q. 

18. Simple Compression. — If the straining action on a bar is an 
axial thrust it produces a longitudinal compression and a lateral 
expansion. The intensity of the pressure on a normal cross section 



of area A with a load P will be f. 



-T- . If we call c the lonoji- 



tudinal compression per unit of lengih we have similarly to equation 
(6) 

fc 



c = 



E 



(9) 



In general, the value of E for compression is the same as for 
tension. 

When the bar is more than about five diameters in length, the 
tendency to bend, or buckle, must be taken into consideration, and 
the strength then determined by the laws for long columns. 

19. Thin Cylinders under Internal Pressure. — Let the figure rep- 
resent a thin cylindrical shell of length 1 and diameter d and let t 




Fig. 8. 

be the thickness of the metal of the plates, all in inches. Let the 
internal pressure be p pounds per square inch. Now suppose the 
cylinder to be cut by a diametrical plane abed; then the total 
pressure or force acting on either side of the plane is p x area abed. 
Call this force P; then P = pxlxd. This force tends to tear the 
cylinder apart along the lines be and ad, and produces a tensile 
stress in the metal the intensitv of which we will call f. The area 



38 ENGINEERING MECHANICS. 

over which this stress is distributed is be x thickness + ad x thickness, 
or is 21 X t, and, consequently, the total resisting force is 2ltf . Put- 
ting the total load and the total resistance equal to each other we 
have pld = 2ltf ; so that the circumferential stress is 

f=g. (10) 

Xow consider the total bursting force as acting axially, then the 
shell tends to tear apart around a circumference as at efgh. 




Fig. 9. 

7rd^ 

The total load P is now — — Xp, and the area of metal section 

resisting rupture is Trd X t. Let f , as before, represent the intensity 
of stress produced. Then equating load and resistance we have 

^ p = 7rdtf, and the longitudinal stress is 

f=M. (11) 

Whence it is seen that the longitudinal stress is half the circum- 
ferential stress. 

Formula (11) is also applicable to a spherical vessel under an 
internal pressure. 

The above formulae are used in calculating the thickness of boiler 
shells. Such shells are made up of several plates riveted together, 
having both longitudinal and circumferential seams. If the whole 
shell were made of a single homogeneous plate without seams we 

would have, from formula (10) i— ^. , but in the actual case the 

efficiency of the riveting must be considered. Call this efficiency t). 
Then 

t=g. (13) 



resista:n'ce of materials to straining actions. 39 

Formulffi (10), (11) and (12) are used for thin cylinders, where 
the thickness t is ver}' small compared to the diameter d. 

20. Thick Cylinders. — When the thickness of the c^dinder is an 
appreciable proportion of the internal diameter, as in the case of 
hydraulic cylinders, the mean stress is unaltered, but the inner 
layers of the metal are more severely stressed than the outer. In 
such cases one of the following formulae are used : 

t=4(^f±P_l) Lame (13) 

In these formulae, t = thickness in inches, d = internal diameter in 
inches, f = the working stress allowed and p = the internal pressure 
in pounds per square inch. In designing such a cylinder it is usual 
to allow for a test pressure of double the ordinary working pressure, 
and to use a value of f = 9000 for cast iron, or 25,000 for steel cast- 
ing. That is, these values of stress must not be exceeded under the 
test pressure. 

(For the deduction of these formulae the student is referred to 
any complete work on mechanics.) 

The fact that the inner layers of thick cylinders are under greater 
stress than the outer, has led to the practical device of making the 
cylinder of several concentric tubes, shrunk over each other in such 
a manner that the inner tube is under an initial compression when 
the pressure is not acting. Familiar examples of this method are 
the built-up gun, used in the service, and hydraulic cylinders. The 
outer tube is bored to a slightly less diameter than the outside of the 
tube over which it is to be placed ; it is then heated until the expan- 
sion is sufficient to aliovr it to be slipped over the inner tube. On 
cooling, it produces a compression of the inner tube, and a tension 
in itself. If, now, a pressure is applied inside the c^dinder, the first 
effect is to overcome the initial compression of the inner tube and 
to increase the tension of the outer: so that the resulting stress in 
the inner tube is the difference of the initial compression and the 
working tension, and in the outer tube the tension produced is the 
sum of the initial and that due to the working pressure. This dis- 
tributes the stress more equally throughout the w^alls of the cylinder. 

21. Resistance to Shearing. — A shearing action is one which 
causes sliding parallel to the section considered. Thus, in Fig. 10, 
the action of the blades of a shearino- machine are shown. The 



40 



ENGINEERING MECHANICS. 



pressure causes a shearing stress in the plane ab. The intensity of 
the shearing stress is the pressure P divided b}^ the cross-sectional 
area A, and is denoted by is- Thus 

f.=:|-. (15) 

rig. 11 illustrates the action wlien the force and reaction do not 
act in the plane of the section. There is now a distinct bending 






] [ 



a 



/ 



Fig. 10. 






Fig. 11. 



action produced, and the effect is to alter the distribution of the 
stress over the section. At a plane half-way between the upper and 
lower surfaces the shearing is greater than the average, while at the 
upper and lower edges of the section there is no shearing action. 





( o 


\ 






\ 


b—G 


P— 


— »- 


1 * ^ 






1 ' 


V- 








t 


i] 






Fig. 12. 



Fig. 13. 



Well-driven rivets are in shear, and well-fitted bolts are usually 
considered to be in shear. On the other hand, pins and cotters are 
rarely in pure shear, in practice. 

Figs. 12 and 13 represent the action in the cases of well-fitting 
and loose rivets. In Fig. 12 it will be seen that the forces really 
act along the centers of thickness of the plates, but, on account of 
the friction and rigidity of the edges ab and cd, the action of the 
forces is practically across the section be. In practice, it is assumed 
that rivets are subjected to pure shearing and their size is calculated 



RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



41 



from equation (15). In the case of pins and cotters it is nsnal to 
calculate the strength on the assumption that they are subject to 
bending. 

Unwin gives the following values of the maximum stress, based 
on experiment^ for such cases : 

P . 

fg = 1.5 — if the section is rectangular and P perpendicular to 

one side. 



= 1.33 -- if the section is circular or elliptical. 

p 

= 1.59 -T- at half-way between the angle and center, if the sec- 

A. 

tion is square, and P acts parallel to a diagonal. 

P . 

= 2 — - if the section is a ring of small thickness compared to the 

diameter. 
22. The Nature of Shearing Stress."^' — Consider an originally 
square particle of thickness unity, to be acted upon by forces P 



-Z' 










d 
i 


t 






-^A^ 


__ 




1' 


a 




d 


I ■ 


^1 








I 


' 


d 




c 


1 
I 






~^ 


~ 















Fig. 14. 



Pig. 15. 



parallel to two of its opposite faces. The square will be distorted 

p 

into a rhombus, Fig. 14, and the shearing stress will be is— -y (^^^ 

thickness being unity). The particle would now spin around due 
to the couple P x ad, or P x be, unless an equal and opposite couple 



* From Goodman. 



42 



ENGINEER] 2sTG MECHANICS. 



is applied to it. In order to make the following remarks perfectly 
general, we will take a rectangular plate, as shown in Fig. 15. 

The plate is acted upon by a clockwise couple, P x ad, or is X ab 
xad, and a counter clockwise conple, P-^xab, or fs'xadxab, but, 
since the plate remains in equilibrinm, these must be equal; then 
fsXabxad = f/xadxab, or f,s = f./; i. e., the intensity of the stress 
on the two sides of the plate is the same. In other words, when a 
force acts so as to produce a shearing stress on opposite sides of a 
particle, there is a stress of equal intensity induced on the other 
two faces, at right angles to the first. 

Xow^ for convenience, we will return to our square particle. The 
forces P and P^ acting on the two sides may be resolved into forces 
E and E., acting along the diagonals as shown in Fig. 16. The 




Fig. 16. 



Fig. 17. 



effect of these forces will be to distort the square into a rhombus 
exactly as before. 

XoTE. — The rhombus in Fig. 14 is drawn in a wrong position 
for simplicity. 

These two forces act at right angles to one another; hence we see 
that a shear stress consists of two equal and opposite stresses, a 
tension and a compression, acting at right angles to one another. 

In Fig. 17 it will be seen that there is a tensile stress 'acting nor- 
mal to one diagonal, and a compressive stress normal to the other. 
The one set of resultants, E, tend to pull the two triangles abc, acd 
apart, and the other resultants to push tlie two triangles abd, bdc 
togther. 

Let fo = the stress normal to the diagonal. 

Then f o X ac = f o X \/2ab = f „ x V2bc = E. 

But V"2xP=V2fsXab=V2fsXbc = E. 

Hence io — i^^ — ij. 



RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



43 



'7 



Thus the intensity of stress is equal on all the four edges and on 
the two diagonals of a rectangular particle subjected to shear. 

23. Modulus of Transverse Elasticity^ or Modulus of Rigidity. — 

Fig. 18 represents an originally 

square particle abed distorted into a o^ I? If* 

the rhombus a'b'cd under the action I 1 Tl /i 

of a shearing force. The amount of 

distortion aa' or bb' is called the 

slide; let this equal x. Then, calling 

1 the original length of a face, the 

ratio x/1 is the distortion per unit of 

lengthy and is called the sliding. Let 

n = x/l, then G-^ the modulus of rig- 

.,., stress J. , 
idity = - — r- = is/n. 
^ strain 

From the figure it is seen that n is the tangent of the angle of 
distortion 0. When the limit of elasticity is not passed, the defor- 
mation is small and we then have 

6l = tan(9 = n 
and 



i7 a' 


/> 


1 




1 




1 




1 




1 




1 N 
/ 
/ 


H- 




/ 




/ 




/ 
1 


/ 



Fig. 18. 



G = f./n=:f,/^, 



or 



(6 being expressed in 
radians). 

The relation between 
the direct elasticity E 
and the transverse 
elasticity G depends 
on Poisson^s ratio, 1/m 
(see Art. 17). 
mE 



*G = 



E = 



2(m + l) 
2(m + l) 



m 



0, 




The averacre value of m beino: 



we have 



G = AE 



* This relation is obtained as follows: It has been shown (Art. 23, 
Fig. 17) that a shearing action may be resolved into a tension Rj, act- 
ing along a diagonal, and a thrust R, perpendicular to it, producing 



44 engineering mechanics. 

Questions and Problems. 

Define resilience and the modulus of resilience. 

Discuss the effect of sudden and impulsive loads, showing how to 
find the stresses produced b}^ a load applied suddenty or impulsively 
compared to that produced b}^ the same steady load. 

Discuss simple tension (or compression) and show why, or- 
dinarily, the stress on a normal cross section only need be consid- 
ered. Show how to find the value of the normal and tangential 
stresses on any oblique section of a material under simple tension 
(or compression). What is Poisson's ratio? 

Deduce equations for finding the thickness of boiler shells to 
withstand a given internal pressure p. Explain the reasons for 
building up thick cylinders on the principle of initial compression. 

What is shearing? Explain the internal stresses produced when 
a material is subject to shearing forces. Define modulus of rigidity. 

equal stresses, to the stress along tlie faces of the particle. If the face 
dc is supposed to be rigidly fixed, the square particle abed takes the 
form of the rhombus a'b'cd under the shearing action. 

The elongation of db due to Ri == f s X db/E. 

The elongation of db due to R = fg X db/E X m (see Art. 17). The 
total elongation of db is then 

kXdb/^_j_J^\^^^ db^rabX V^. 
E V m / 

Now, for simplicity, consider the original length of each face of the 

particle = unity, then bd ■=^ V^ and we have 

elongation =x= ^ X v2" 1 1+ ^j . (a) 

Draw be perpendicular to db', and remembering that (? is a very 
small angle, we have 

eb' = elongation of the diagonal = x. 
X = db' — de = db' — dcV'^^ db' — V"2. 
From trigonometry c = a sio c cosec A, = 180° — (A + B) 

sin C = sin (A + B) = sin ^ 1 -^ ~~ "9 ) =^ ^^° ( "T ~ 'o J ^°^ (t ~ Y J 

||co^ 

:e*dc=l. 

I ^ ^ \ 7T, 0/ » d . TT . d \ yr-, 

xr=2 COS X ~ o^ — >/'^ = '^ I cos^ COS g- + Sin J- sin ^ I — v ^ 



.. c = db' = 2dc '■ !^'"1^ '! =8dc cos f^ - -I) 
sm ' — ' • ' 



= 2cos^^ 



since*dc=l. 



resistance of materials to straixixg actions. 45 

Probleinis. 

1. Find the stress and the extension produced in a bar 10 feet 
long and l-J square inches section, by the sudden application of a 
tensile load of 6 tons. What suddenly applied load would produce 
an extension of 2V i^^^-l^? E = 26,000,000. 

2. A load of 500 pounds falls through ^ inch onto a stop at the 
end of a vertical bar 10 feet long and 1 square inch section. Find 
the stress produced in the bar. E = 26,000,000. 

3. Find the greatest height from which the load of the above 
problem could fall, before bringing up on the stop, so as to produce 
a stress not greater than 11 tons per square inch. 

4. A cylindrical vessel with hemispherical ends, diameter 6 feet, 
carries a pressure of 200 pounds per square inch above the atmos- 
phere. The cylindrical part is constructed of steel rings, riveted 
together. If f=7 tons per square inch, how thick must the metal 
be, and what is the longitudinal stress in the metal of the ring joint 
whose section is -^ that of the solid plate. 

5. A round iron 1 inch diameter sustains a tension load of 20,000 
pounds; find the unit normal and tangential stresses on a section 
at an angle of 30° with the normal section. Solve also for a section 
at 45°. ^ 

6. From the values of E and G given in the table. Art. 15, find 
the value of m for steel. 

7. An eye bar carries a steady load of 15 tons. What must be 
the diameter of the wrought-iron pin to resist shearing. 

XoTE. — Consider the pin in double shear, and assume this shear 
uniformly distributed over the area of the pin. Use Table 8. 



But ,y being very small sin -^ = ^ and cos -^ =1. 

f, f, 

Bv definition G— ~ , ft- p^ ■ 
u (j 



x=^^=™~^2(l+ ^ from (a). 



1 2 / m+1 

G" = E I ~m" 

P Em 

^- 2(m + l] 



CHAPTER IV. 

IiESISTANCE OF MATERIALS TO StEAINING ACTIONS (CONTINUE!)). 

BENDING^ LIMITATION OF THE THEORY OF SIMPLE BENDING, MOMENT 
OF INERTIA, MODULUS OF SECTION. TABLE OF I, Z, ETC. CURVES 
OF SHEARING FORCE AND B. M. TABLE OF S. F., B. M. AND 
DEFLECTION DIAGRAMS OF S. F. AND B. M. 

24. Bending. — For the general tlieor}^ of beams^ see Smith, 
" Strength of Material." 

In order to have simple bending the following conditions mnst he 
fulfilled: (1) The axis of the bar is a straight line, joining the 




Y^ 



Fig. 20. 

centers of hgure of pai'allel transverse sections; (2) the bar is sym- 
metrical about a plane passing through the axis; (3) all the external 
forces act in such a plane of symmetry, calknl the plane of bending, 
normally to the axis (Unwin). 

Fig. 20 represents a beam under simple bending, due to the action 
of the forces, as sliown in the iigiire, the flexure being greatly 
ex«aggerated in the lower figure. 




RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 47 

In order to obtain a starting point for 
the consideration of bending actions, cer- 
tain assumptions must be made, as fol- 
lows : 

(1) Plane transverse sections remain 
plane and normal to tlie longitudinal axis 
after bending. 

(2) The material is homogeneous, and 
obeys Hooke's law, and the strains pro- 
duced do not exceed the elastic limit. 

(3) Every layer of the material is free 
to extend or contract under stress, as if 
separate from other layers. 

(4) The modulus of elasticity is the 
same for tension and compression. 

Returning now to Fig. 20, since the strain- Fig. 21. 

ing action is the same on every, section be- 
tween A and B, the curvature of tlie beam will be circular. Con- 
sider any two transverse sections, CF and DE, close together ; after 
bending they are no longer parallel, but the layer of material CI) 
is stretched to CD', and the layer EF is compressed to E'F', while 
the layer HC is neither stretched nor compressed. The surface, 
then, throuoh HG is not under anv lono'itudinal stress, and is called 
the neutral surface, and its line of intersection XX with a transverse 
section is called the neutral axis of the section. 

Suppose the sections C'F' and D'E' produced to meet in a line, 
which being seen end on, is represented by the point 0, Fig. 21, the 
angle of intersection being 0. Call the raditts of curvature of the 
neutral surface E, and let y be the distance from WG' of a layer 
I'K', originally parallel to the neutral surface. Then 

FK' _ (Pv+y)^ _ R + y 
H'G' E^ R • 

The strain of the layer I'C is then 

_ FK^-IK _ FK'-H^G' _ (E + v)^-E^ ^ y_ 
^- IK H'G' E^ R • 

The intensity of the stress thus produced, f, being by assumption 
within the elastic limit, is, then, 

f = Exe = E ^-(equation 6). (15) 



48 



ENGINEERING MECHANICS. 



The intensity of the compressive stress will have the same value, 
since by assumption E has the same value for compression as for 
tension. 

The intensity of the longitudinal stress at every point in the 
cross section is, then^ proportional to its distance from the neutral 
axis. 



. —^i^^LfiL^C^jflce^ F' 




Fig. 22. 



Fig. 23. 



Thus^ at the neutral axis (y = 0) the stress is zero^, at unit dis- 

E 

tance from the axis (y = l) its value is ^ , and it is a maximum at 

the boundary furthest from the neutral axis. 

The variation of the longitudinal stress acting on any section CF 
is shown by Fig. 22. These stresses have resultants FF, which form 
a couple Fxd, as shown by Fig. 23. This may be called the couple 
of the internal stresses. 



•§^ 




«? 



oc 



Be 



Fig. 24. 



Again, returning to Fig. 20, and considering the external forces 
acting to the left of the section CF, the bending moment is seen to 
be Pl^ — Pig. Since equilibrium is sustained, this bending moment 
is balanced by the couple of the internal stresses, or 

M = P(li-l2)=Fd. (16) 



EESISTANCE OF MATERIALS TO STEAINING ACTIONS. 49 

In Fig. 20, the section of the beam is shown symmetrical about a 
horizontal axis^ but this is not necessary; the beam may have any 
section symmetrical about the plane of bending YY as shown in 
Fig. 24. 

The algebraic snm^ of all the external moments acting must be 
zero, since the beam is in equilibrium. Also the total of the tensile 
forces must balance the total of the compressive forces, since they 
form a couple balancing the total of the external forces. Therefore, 
the algebraic sum of the internal horizontal forces is equal to zero. 

In Fig. 22, let 8 A, or xdy, be an elementary strip of area, parallel 
to the neutral axis, x being the variable width of the strip. The 

stress on this element is f x8A=: — ^ — (from equation 15) and the 

total stress on the section is 

y Ey^A 

but since the total stress is zero 

F 
and, since ^j^- is a constant, 

^y8A = 0. 

But this relation can hold only if the distances y are measured from 
an axis passing through the center of gravity of the cross section. 
Therefore, the neutral axis passes through the center of gravity of 
the section. 

The moment of the stress acting on the strip of area is f8x4y 

= ^ . The moment of the total stress acting on the section 

is then 

^/Ey25A\ E ^ 2^, EI 

1 being the moment of inertia of the cross section. This total 
moment of the internal stresses is called the moment of resistance 
of the section. 

If, now, we equate the external bending moment M to the moment 
of resistance of the section, we have 

M=f, (16) 



50 ENGIMEERING MECHANICS. 

expressing the relation between the bending moment and the curva- 
ture of the beam. 

If now f is the stress at a distance y from the neutral axis we 
have from equation 15 

. Ey E f 

^= R "'^ R=T' 
Equation (16) then takes the form 

R y 

whence 

f = M-|-. (17) 

Equation (1?) enables us to obtain the stress at any point of the 
section of a beam distant y from the neutral axis^, knowing the 
value of the external bending moment^ and the moment of inertia 
of the section. Since it has been shown above that the greatest 
stress occurs in the fibers most distant from the neutral axis, the 
problem is, generally, to find the greatest tensile or compressive 
stresses produced, and to so fix the dimensions of the section that 
these greatest stresses do not exceed the safe working stress allow- 
able under given conditions. Let y^ be the distance of the most dis- 
tant fiber in tension, from the neutral axis, and jc the distance of 
the most distant fiber in compression. Then the greatest tensile 
and compressive stresses are 



f, = M ^ andf, = M^ 



Let 



— =Z/^and — =Zc, 
Jt Jc 

then 

M = f,Z, = f.Z,. (18) 

Zt and Zc are called the moduli of the section with respect to ten- 
sion and compression. If the section is symmetrical about the 
neutral axis, jt = Jc and Zt = Zc, and it is then necessary to consider 
only the stress, whether tension or compression, that the material 
is weakest to resist. 

25. Limitation of the Theory of Simple Bending. — The theory of 
simple tending explained in the preceding article refers only to 



RESISTAXCE OF MATERIALS TO STRAINING ACTIONS. 51 

cases in which there is no shearing force, but in many cases shearing 
force is present as well as bending. In such cases the internal 
stresses acting at any section have not only to balance the bending 
moment, but also the shearing force, and, consequently, there will 
be tangential as well as direct longitudinal stresses acting on any 
section. The case, then, becomes one of compound stresses, which 
will be investigated later. 

However, in most practical cases, the theory of simple bending 
is sufficient to enable the calculation of stresses and strains to be 
made with a workable degree of approximation. In many such 
cases the greatest bending moment occurs at sections across which 
the shearing force is zero, or negligible, and very frequently, where 
the shearing force is greatest, the bending moment is small enough 
to be neglected. The usual practice of the engineer is to follow the 
simple bending theor}^, allowing for doubtful cases by a proper selec- 
tion of the allowable safe working stress — in other words — by a 
judicious selection of the factor of safety, being guided by previous 
successful solutions of similar problems. 

26. Moments of Inertia. — Properly speaking, the moment of 
inertia of a body about a given point, or line, is the sum of the 
products of the elementar}^ masses of which it is composed by the 
squares of their distances from the point or line. This definition is 
extended in meaning when applied to plane surfaces as the sum of 
the products of the elements of area by the squares of their distances 
from the point or line. Thus, let dA be an element of area at a 
distance y from the axis; then the moment of inertia of the sur- 
face is 

l = MAf. 

If A is the total area of a surface and I its moment of inertia about 
an axis through the center of gravity, then 

I/A=k^ 

and k is called the radius of gyration of the section. 

In order to facilitate the use of formula (18), M = fZ, all text 
books,. hand books and reference books for the use of engineers con- 
tain tables showing the values of moments of inertia, radii of gyra- 
tion, moduli of sections, etc. Such a table, for a few of the simpler 
sections, is given in Table 9, Art. 27. 



52 



ENGINEERING MECHANICS. 



The values given in such tables are necessarily based on a given 
axis, usually the neutral axis of the section^ but it is frequently 
desirable to have the value of the moment of inertia about some 
other axis. 

27. Moment of Inertia about Parallel Axes. — In Fig. 25 suppose 
it is wished to obtain the moment of inertia of the area about the 
axis ZZ. Let I be the moment of inertia of the surface about an 
axis X'X through its center of figure obtained from the table. Let 
Ii be the moment of inertia about the new axis ZZ^ parallel to XX, 




Pig. 25. 



and distant Y from it. Let 8A be an element of area, distant y 
from the axis XX, and let A be the total area of the surface. Then 
I = ^8Ay^ 

Ii = :§8A(Y + y)2 = :^8Ay2 + 2Y^8Ay + Y228A, 
but 

S8Ay = 0, 
since y is measured from an axis passing through the center of 
figure, and 

28A = A, 



.•.I, = I + AY^ 



(19) 



Equation (19) is used for finding the moment of inertia of a 
complex section which can be divided up into a number of rec- 
tangles or other simple sections. Thus, suppose we have a beam 
of I section and wish to find the position of the neutral axis and 
moment of inertia of the complete section. Divide the section into 
rectangles, as in Fig. 26. To find the distance of the neutral axis 



EESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



53 



from some fixed line, as^ for instance, the base line of the figure 
XX, take moments about this line^ thus : 

Moment of upper flange = Axa. 
Moment of web flange = B X b. 
Moment of lower flanges Cxc. 

Total moment = Axa + Bxb + Cxc. 
Total area =:A + B + C. 

_ equation (a) 
~ equation (b) * 

which locates the neutral axis xx of the combined section. Now, 
calling the distances of the neutral axes of each part considered 



Then, the arm, 



(a) 
(b) 

(c) 

























^ 






^ 


\ 




B 




; 


?! 


JC 










X 




-< 


i 






t 


^ 


.^ 














^ 








c 


X 



Fig. 26. 

separatel}^ from the combined neutral axis, y^, Jb.Jc, 
ment of inertia of each part, \a, Ib ; 
tion (19) 

Itotai = I^ 4- Ib + Ic + Ay2^ + By^B + Cy^c 



and the mo- 
le, we have, from equa- 

(20) 



The value of \toiai having been found thus, it is divided by the 
distance of the most distant fiber in tension, and in compression, 
thus giving the value of Zt or Zc and these values of Z are used in 
equation (18). 

The axis of moments is the horizontal broken line through the 
center of figure of the section. I is the moment of inertia about the 
axis through the center of figure and at right angles to the plane of 



54 ENGINEERING MECHANICS. 

TABLE 9.— PROPERTIES OF SOME SIMPLE SECTIONS. 



Shape of Section. 




Vi>^ 



. Moment of 

^^^^^ Inertia. 

Section. j 



^ S4 

13 ^ 



W7. 




bh 



. bh3 



Square of 

Kadius 

of Gyration. 



12 ^ 



i- 





bh 



36 



bh^ 



18 



h2 



-^^D^==.0491D^ I ^D-^ 




( D4-d^) 



Modulus 
of Section. 

i/y 



S3 



bh- 



32 



D=^=.0982D3 



^MD-^+d^) j 



If D— d IS small ^ y p ^ 



1^ D- nearly 



-." ba=^ 



5o ba2 



EESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



55 



bending. Where two values of Z ai-e given^ Z^ is the modulus for 
fibers at top edge of figure, and Zj for fibers at bottom edge. 

28. Beam Fixed at One End. — This case is a little more complex 
than the case of simple bending considered in Art. 24. Suppose a 
force W acts on the end of a beam which is solidly fixed at its other 
end, as in Fig. 27, and that it is required to find the action at the 
section ab. If we introduce two equal and opposite forces W and 
W", each equal to W, at ab, the equilibrium of the beam is not dis- 
turbed. Then the action of W on the section ab is equivalent to 
the couple AVW", and the unbalanced force W. The moment of the 
couple is Wxl and is balanced by the moment of the internal 
stresses in the manner described in Art. 24. The other force W 



W^' 



o\ 



M 



W 

Fia. 27. 



W 



produces shearing stress across the section, and the beam must be 
strong enough to resist both the bending and shearing stresses. If 
there are several forces acting to the right of ab they may be re- 
duced to a single equivalent resultant force. 

In many cases the effect of the bending moment is much greater 
than that of tlie shearing force, and the latter may be neglected. 

29. Curves of Bending Moment and Shearing Force. — Fig. 28 
represents a beam supported at the ends and carrying a distributed 
load of varying intensity. Consider any point C in the beam, dis- 
tant X from A, and let the intensity of loading at the point C be 
w. For any small length dx in the vicinity of C then we have the 
load equal to wdx. Let E be the reaction of the support at A. At 
every point along AB erect an ordinate of length, to some con- 
venient scale, equal to the value of w at that point. Draw a curve 
through the end of these ordinates, and we have the load curve 



56 



ENGINEERING MECHANICS. 



DHE. Then the area ADHEB represents the total load on the 
beam, and the resultant of this load acts through the center of figure 
of the area. The shearing force at C is the resultant of all the 
forces to the left of C; call this shearing force S. Then 

S = E-J^^wdx. (a) 

If the value of S is calculated for a number of points along the 
beam and ordinates erected to represent these values, a curve 
through their ends will give the curve IK, which is the shearing- 
force curve. In the same way, calculate the value of the bending 




Fig. 28. 



moment, M, at C. This moment is the algebraic sum of the moments 
of all forces to the left of C, and is 

M = j^Sdx. (b) 

Construct the bending moment curve, AFB. 

From equations (a) and (b) it is seen that the shearing force at 
any point is equal to E minus (the area of that part of the load 
curve between A and the point) and that the bending moment at 
any point is equal to the area of the shearing-force curve between 
A and the point. Also from the figure, the bending moment is a 
maximum at the point at which the shearing force changes its sig]i. 
A case of frequent occurrence in practice is that in which w is con- 
stant (that is, the load is uniformly distributed), in which case the 



RESISTAXCE OF MATERIALS TO STRAINING ACTIONS. 



57 



curve of shearing force is a straight line, and the curve of bending 
moment is a parabola. 

Table 10 (following) shows the greatest shearing force, bending 
moment, stress and deflection of beams under different loads and 
methods of support, and Table 11 shows shearing force and bending 
moment curves. 

30. Beam Fixed or Encastre. — A beam is said to be fixed, or 
encastre, when its end is so supported that the direction of the axis 
of the beam is not changed by the action of the bending forces. 

This Fig. 29 represents a beam fixed at one end and loaded with, 
a weight W at the other end. 





y? 






^ 


V 






IX 


M 


/ 




^-oc^ 









B-^W 



W 



Fig. 29. 



Such a beam may be considered as under the action of a couple, 

Ex = Wl. 

31. Economy of Different Sections. — The resistance of a beam to 
bending is proportional to its sectional modulus Z. Thus, if we 
have a number of beams of different cross sections, under the same 
loads, all having the same value of Z, that one is most economical of 
material which has the least area. Therefore, the greater the ratio 
of Z to A the more economical is the form of the beam. 

In a beam of circular or rectangular section the top and bottom 
fibers only are fully strained. iVt the neutral axis there is no strain, 
due to bending, so that by removing material from the vicinity of 
the neutral axis and placing it at the top and bottom edges the 
beam is made stronger. This gives the I or double-flanged section. 
In a beam of this section the flanges resist nearly the whole of the 



58 



ENGINEERING MECHANICS. 



stresses due to bending, while the shearing forces are resisted princi- 
pally by the vertical web. 

In a beam of I section, in order that both flanges may carry their 
due proportion of the result of the load, the following relations must 
hold : 

M = f,Z, = fcZo. 

If f^ = fc then Tit^Tic, which is the case when the section is sym- 
metrical about the neutral axis. If, however, f^ does not equal fc, 
the two flanges will not be under the same stress if Z^ = Zc, and, 
therefore, in order to have the most economical distribution of 
material the form of section must be unsymmetrical about the 
neutral axis. 







I 


-i 
























JI 




^ 










^^ 








c 






1 















Fig 30. 



Unsymmetrical Section. — Many materials have very unequal 
strength to resist tension and compression, that is, f* and fc hav3 
different values. 

In Fig. 30 let A = area of the web of a beam, B = area of flange 
in tension, C = area of flange in compression and h=:the depth of 
the beam, from center to center of the flanges. Suppose the area 
A of the web is small compared to the areas of the flanges. Then, 
in order that each flange may be stressed in proportion to its 



strength, we have 



ffXB = feXC. 



(a) 



Now, let the neutral axis of the complete section be at a distance 
he from the center of the lower flange. Assuming that the effect of 
the web in resisting Ij ending is negligible, and calling M the ex- 



RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 59 

ternal bending moment, we have^, taking moments about the neutral 
axis, 

M = f,B(ll-h,)+feChe. 

Substituting from (a) 

M = f,B(li-h,)+f,Bxhc, 
.•.M = f3h 



or 



similarly, 



^= UK-' (- 



C=g. (22, 



If S = the total shearing force and fs the safe shearing stress of 
the material the necessary area A of the web is given by 

A=|-. (23) 

The assumption that the web does not resist any of the bending 
action may, in some cases, especially when the web is heavy, intro- 
duce an error, but this error is on the safe side. In the usual cases 
of rolled steel sections the error is inappreciable. 



60 



ENGINEERING MECHANICS. 









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© © 








a. 


P.^ 








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tB+3 








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CI 

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S3 . 

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la 


p5 


M 


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C! 


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«H 


a 


a^ 


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4^ 


c. 




03 




S^ 


o-g-o 




Id 


9 


^ s 


i^c^ 




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o 


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H^ 


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tD 


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ai» 


^ « ^ 




CO 'l— 1 


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^te 


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-^J 


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ormal 
ss due 
load, 
fb 


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tS] 


lb 
















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f 






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^l-ll 


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sitio 

of 

3ates 

ndin 

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plh bii^a 


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k^wwv 


N.<\\\\V; 


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RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



61 



o 

.513 

P n< . 

SS O t» 












^u 


a 


"(islH 


4^ 


^|S 



^ 












'sS "' 


'^ 1 


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"cs!" 


c i2. c; 


^,H 






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•a .5^ 




62 



ENGINEERING MECHANICS. 



< 


H 


Z 

K 




|-) 


W 


H 




H 






o 


O 


£ 




^ 


H 






O 


u 


o 


;?; 


w 


iz; 


' ' 


X 






S MO 



(D c S 
J^ o 



^|H 



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2i. S3 











; -Q 






'o 


c5 


2.^^ 


oi 




"fe 


^^1 


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RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



63 



TABLE 11.— DIAGRAMS OF SHEARING FORCE AND BENDING 

MOMENTS. 



Loading-. 



Diagram of load. Shearing Force 
Curve. Bending Moment Curve. 



Fixed at one 
end. 

Single load 
at free end. 



Remarks. 




The only moment acting to 
the right of A is Wl, which 
is therefore the bending 
moment at A. 



This is simply an example of 
the combination of two 
diagrams such as case 1. 



Fixed at one 
end. 

Uniform 
load of w 
lbs. per 
inch run. 



The load being uniformly 
distributed, its resultant 
acts at i 1 from A. The total 
load is wl or Q. .*. the 
bending moment at A=wl 



64 



ENGINEERING MECHANICS. 
TABLE 11 (Continued). 



Loading. 



Diagram of load. Shearing Eorce 
Curve. Bending Moment Curve. 



Supported at 
both ends. 

Single cen- 
tral load. 




Supported at 
both ends. 

Single load 
not central. 




Eemarks. 



Each 



load = 



support 
W 



takes 



the 



The only mo- 



ment to the right or left of 

W 1 Wl 
C isi-Y-X~2=~T'- ^tany 
section the moment varies 
directly as its distance 
from the support, hence 
the diagram is triangular. 
The shearing curve changes 
sign on opposite sides of 
the middle section. 



Taking moments about one 
support gives 

Ell = WloIll=^^^^ 

The B. M. at C is 

B.M. =Il,l, = ^\ 



Supported at 
both ends. 

Two sym- 
metrical 
equal loads. 




^.M./fi./7^^ /iGrncf loa d 






The sum of the forces to the 
right or left of a= W— R=0, 
and to the right of D. or to 
theleftofCthesum=R=W. 

The only moment to the 
right of D is AVli, similarly 
to the left of C. At any 
section x between the loads 
and distant \x from one of 
them, we have, by taking 
moments to the left of x, 

R(li + 1^) — Wlx 

= R1] + R1^ — Wla; 

= Rli or Wli. 



RESISTANCE OF MATERIALS TO STRAINING ACTIONS. 



65 



TABLE 11 (Continued). 



Loading. 



Supported at 
both ends. 

Qniformly 
distributed 
load, w lbs. 
per inch 
run. 



Diagram of load. Shearing Force 
Curve. Bending Moment Curve. 




Remarks. 



Supported at 
tAvo points 
equidistant 
from ends, 
and uni- 
formly- 
loaded with 
w lbs. per 
inch run. 



^. M. Co/nStnec/. 



Replace the forces by their 
resultant. Each support 

wl 
takesone half the load =~n • 

The forces acting to the 

right of C = ^ — ^- = 0, or 

shear at middle section in 
zero. 
Taking moments to left of C 
gives 

wl_ J__ wl J_ _ wl2 
32 2^4"~8" 

wl 
The -g" Shown midway be- 
tween C and the support is 
the resultant of the load on 
half the beam acting at its 

center of gravity, -r- from 

C, or from the support. At 
any other section x distant 
\x from the support, find 
the B. M. as follows: Take 
moments to the right of x, 

^Xlx-Wlx x-^ 

= :^ (1-1«) (a), 

let 1— lx = l'x, then (a) be- 
comes 



(IxXl'x) 



_ Q 



Q l \x Vx \ 

rv 2 ) 



the curve is a parabola. 



The shearing force curve is 
a combination of cases 3 
and 7. The bending mo- 
ment is a combination of 
cases 3 and 6. This case 
illustrates the importance 
of giving moments their 
proper signs. The B.M.Avill 
be smallest when Ma (or Mb) 
= Mx, or when 

wlj _ wi; _ wlj 
2 8 2 



wl- 



wlf 



1? 



=8, li=2.83l2. 



But li + 2I2 = 1, whence 
2.83l2 + 2l2 = li or 1 = 4.8310, 
or say 12=51 for maximum 
strength of beam. 



66 



EISTGINEERIJTG MECHANICS. 
TABLE 11 (Continued). 



Loading. 



Diagram of load. Shearing Force 
Curve. Bending Moment Curve. 




Remarks. 



For method of this case see 
Goodman, Machines Applied 
to Engineering^ or Smith, 
Strength nf Material. 



See Goodman or Smith. 



Questions and Problems. 

Discus? the conditions of simple bending. Show liow to deduce 
the value of the stress f in a beam, in terms of the bending moment 
M and the m.odnlns of the section, Z. 

Explain the limitation of the theory of simple bending. Wh}' 
may it safely be used in most practical cases? 

Define moment of inertia. Given the moment of inertia, I, about 
an axis through tlie center of figure of a plane area, show how to find 



CHAPTER V. 

Directions for Practical Work, 
practical problem i. 

32. Directions for Practical Work. — Data differing from that 
IN the text will be given for the problems. 

Calculate problems on loose paper to nearest hnndredth reducing 
this to nearest 3 2d from table in back of book. If the result falls 
midway between thirty seconds in table, take the highest 32d. The 
gauge for accuracy will be the result obtained in hundredths. 
Note that when results are reduced to nearest 32ds these are used in 
succeeding steps of problem. Make smooth copies of calculations 
on interleaved pages. Place important headings in the margin at 
the left of the page. Make a drawing in pencil of the required 
views of the problems on cross-section paper (or drawing paper if 
used) placing all dimensions properly; complete the drawing and 
make a tracing and two blue prints if there is time. Use a whole 
sheet of cross-section paper for each problem. 

Place the legend, number of views, etc., of each problem accord- 
ing to instructions. 

Omit cutting, border and working lines when cross-section paper 
is used and turn in the sheet without trimming it. 

Trim the tracing to the exact size of sheet of cross-section paper. 

If drawing paper should be used, secure the sheet on the boards 
with thumb tacks, and draw cutting, border and working lines of 
the following dimensions, respectively, 18"x24:", 16"x22" and 
15"x2r. 

Each midshipman will use the data corresponding to his own 
desk number. 

Make all lettering, dimension figures, conventional lines, etc., 
conform to the standards of the drawing course. 



DIRIiCTIONS FOR PRACTICAL WORK. 69 



70 



ENGINEERING MECHANICS. 

Problem I.* 




33. Design a knuckle joint for a stay rod of a boiler. Area sup- 
ported 12" X 12". Pressure per gauge, 150 pounds. Material 
wrought iron. tt = '7200 pounds per square inch. 



DRAW FRONT AND SIDE ELEVATIONS t SCALE, FULL SIZE. 

To find the diameter of the rod A, which is in tension, we 



have 



P = 12x12x150 = 21,600 pounds. Area of rod= ^^^ =3.00 



square inches. 

* From " Notes on Machine Design." 



DIRECTIONS FOR PRACTICAL WORK. 71 

And A, the diameter, = 1"95, or l{f\ 

Let B be 1.1 A (Unwin)^ or, 

B = l.lA = l.lxliF = 2"13, or 24". 

To find the diameter of the pin C, which is calculated to resist 
bending. The pin is in the condition of a nniformly loaded beam 
24" long, supported at each end. We have, therefore, 

^, wP , P XT ^/r PI 21,600 X2V' 

M = g- and w= . . Hence, M = g = ^8 — 

^ 21^00X17 

Also, M = ffZ where z= ^^ =.0982 d^ and fi = 7200 pounds per 
square inch. 

Therefore, ^^^^^^^^ := 7200 x. 0982 d^ or. 
0X0 



^y 



21,600X17 ^,.01, or 2' 



8x8x7200x.0982 



Then C, or diameter of the pin, is 2". 

To find D, the thickness of the part surrounding the pin. The 
metal is in the condition of a uniformly loaded beam fixed at both 
ends. 

Fig. 32 shows a perspective view of the eye of the bottom rod and 
of the pin through it. It is understood that the eye is pulling down 
upon the pin. The eye Avill not fail by shearing, as might be as- 
sumed from the figure. This figure is only intended to show the 
portion of metal considered as a beam. Two planes shown by dotted 
rectangles form the ends of the beam where it is " fixed '^ to the 
remaining part of the eye. The little beam is drawn separately 
above, the " fixed '' planes at the ends being shown by ruling them 
with vertical lines. 

1 = lengths C = 2". M= ^ = 21^00x2 ^^qqq,, 

IZ ±/i 

* This is the bending moment at ends of a uniformly loaded, fixed 
beam. 



72 



ENGINEERING MECHANICS. 



Also, M = ffZ where f^r=7200 and z 



bh^ 



b = 2r. 



Therefore, M 



7200xh2xl7 



6X 



= 3600, and 



h=iri9, orD = l-rV'. 



The depth E is taken eqnal to D 
for symmetry and is thus made 
amply strong to resist irregular 
stresses and fitting. 

As possibly two-thirds of the stress 
may come on each arm of the split 
end, due to irregular fitting, F is 
made about f of 6 = 1^42, or lif"- 

As one dimension of the cross 
section at F is the same as one side 
of the square section at B, the other 
dimension alone takes the entire 
variation. 

The lower portions of the forked 
ends are made a little wider than at 
F to make a good bearing for the 
pin, or l-J". Or, from Unwin, 
.75 X diameter of pin^l^". 
Gr, the thickness of the head of the pin, is taken as .5 C (Unwin). 
K, the diameter of the head of the pin, is 1.5 = 3". 
H, the collar, has the same dimensions as the head of the pin. 




Fig. 32. 



J is taken long enough for a round split pin of a diameter = -j- 

and an equal amount beyond. This is empirical. 

The space above D is taken a little greater than D, or 1-J". 
The eight-sided portions at top and bottom are empirical and repre- 
sent the corners of the square head bevelled off to make an eight- 
sided section. The dimensions of this portion depend upon the 
work in the blacksmith shop, as a fuller is used to shape the 
octagon. Make the length that required for use of proper fuller. 

The part where the cylinder merges into the octagonal prism must 
be carefully drawn to show that it is octagonal not hexagonal. 
Draw, in pencil onl}^, a plan like that in Fig. 23, draw the 



DIRECTIONS FOR PRACTICAL WORK. 



73 




^^'^^^ 



octagon and its inscribed and 
circumscribed circles. On the 
front elevation, assume some 
fairly large radius r for the 
arc of enlargement, ab, of the 
cylinder, and draw the arc, ex- 
tending it to g. From g' on 
the plan project down to g on 
this arc. Pass horizontal lines 
through b and g. The lines bf 
and gk represent horizontal 
planes which intersect the sur- 
face of revolution generated by 
the arc abg, in circles. These 
circles are seen in their true 
form on the plan as the in- 
scribed and circumscribed 
circles already drawn. Corre- 
sponding points, therefore, can 
be projected from plan to ele- 
vation as b'b, h^h, c'c, i'i, etc. 
From i and j draw in and jp, 
the near edges of the octagonal 
prism. Find by trial arcs of 
circles to connect hci, idj and 
jek, as shown. 

The fact that b and f are on a level with c, d, e, and are 7iot 
the same points as h and h, shows the difference between octagon 
and hexagon. From n and p any arcs may be drawn to the side 
boundaries. 




Fig. 33. 



^p/if /^g'n 



Fig. 34. 




Fig. 34 shows the split pin. This is made of half-round wire 
bent into an eye as sliown. D, the inside diameter of the eye, = d. 



74 ENGTNEERING MECHANICS. 

d = ^C. After entering the pin the ends are opened out to prevent 
its falling out of place. 

QUESTIOI^S AND PROBLEMS. 

Make a neat sketch of a knuckle joint for a boiler stay, and ex- 
plain clearly the nature of the stresses to which, each part is sub- 
jected and the considerations determining the size of each part to 
resist these stresses. 

A rod of circular section expands into an octagonal section. Make 
a neat sketch on the blackboard showing the construction; plan and 
front elevation. 

Problems. 

1. The stay rods connecting the shell to the front head of a boiler 
are fitted with knuckle joints. The line of the stay makes an anglo 
of 30° with the axis of the boiler. Area of head supported by each 
stay 15"xl2". Pressure per gauge 160 pounds; ff = 7500 pounds. 
Find the diameter of the stay rod, and diameter of the pin. 

I^OTE. — If P is the force acting normally to the head on the area 
supported^ P X sec ^ is the resultant force acting in the direction of 
the axis of the stay. 

2. The opening of the forked end in a knuckle of a tie rod which 
sustains a pull of 24^000 pounds is 2J inches. Find the necessary 
diameter of the pin of the knuckle joint. f^ = 7200 pounds per 
square inch. 

3. In the knuckle joint shown in Fig. 31 find A, B, C and D. 
Area supported^ 14" X 14"^ pressure per gauge 180 pounds ; material, 
wrought iron; fi = 7500 pounds per square inch. 

4. In the knuckle joint shown in Fig. 31 find A, B, C and D. 
Area supported, 10" x 10" ; pressure per gauge 160 pounds ; material 
wrought iron; fi = 7500 pounds per square inch. 

5. In the knuckle joint shown in Fig. 31 find A, B, C and D 
Area supported, 12"xl2"; pressure per gauge 200 pounds; mate 
rial, wrought iron; ff = 7800 pounds per square inch. 



DIRECTIONS FOR PRACTICxiL WORK. 



75 



33. DATA FOR THE DESIGN OF A KNUCKLE JOINT FOR A STAY 
ROD OF A BOILER. 



Problem 
No. 


Desk No. 
ending in 


Boiler 
pressure. 


Area 
supported. 


Material. 


Working strength 
ft- 


1 


1 


180 lbs. p. g. 


14" X 14" 


Wrought iron. 


7,500 lbs. / in. 


2 


2 


160 " " 


10" X 10" 


.. 


7,500 lbs. / in. 


3 


a 


200 " " 


12" X 12" 


" 


7,800 lbs. / in.2 


4 


i 


140 " " 


14"xl4" 




6,800 lbs. / in.2 


5 


5 


180 " " 


12" X 12" 




8,500 lbs./ in.2 


6 


6 


200 " " 


15" X 16" 


Steel. 


9,000 lbs. / in.2 


7 


7 


220 " " 


15" X 15" 


- 


10,000 lbs. / in.2 


8 


8 


150 " " 


" X 11" 


Wrought iron. 


6,500 lbs. / in.2 


9 


9 


200 " " 


15" X 15" 


Steel. 


9,000 lbs. / in.2 


10 





140 " " 


14" X 14" 


Wrought iron. 


6,800 lbs. / in.2 



Use a scale = one-half size for problems 1, 3, 5, 6, 7 and 9 ; for the 
remaining problems use scale = full size. 



CHAPTEE YI. 

Bolts. Nuts. Screws. 

forms of screw threads. applications of bolts, studs, nuts, 
etc. stresses in bolts and screws. friction and effi- 
ciency of screws. calculation of number^ size and pitch 
of cylinder cover studs. 

34. A screw is a cylindrical bar on wHcti is formed a helical pro- 
jection called the thread. The screw fits into a corresponding hol- 
low form, which is called the nut. The mechanical pairs thus 




Fig. 35. — Section of Screw Thread. 

formed are used in the greatest variety of ways^ but these uses may 
be grouped into three general classes : (a) as fastenings, when they 
are called bolts, screws or studs; (b) for adjusting the relative posi- 
tions of two parts; (c) for transmitting power. 

The thread is usually right-handed and should always be made 
so for all purposes, unless, in very special cases, it is necessary to 
use a left-handed thread. 

Bolts and screws used for fastenings are generally subject to ten- 
sion, the forces acting parallel to the axis of the bolt and normally 



BOLTS. :f^UTS. SCREWS. 77 

to the surfaces connected together. When the forces act normally 
to the axis of the bolt and parallel to the surfaces of the parts fas- 
tened together, the bolt is in shear, in the same manner as a rivet. 
The advantage of a bolt over a rivet in such cases is that the con- 
nected parts can be easily taken apart when necessary or convenient. 
In Fig. 35 d is the nominal diameter of the bolt. (This is also 
nsnally the diameter of the plain cylindrical part, or diameter of the 
rod on which the thread is cut.) d^ is the effective diameter, that is, 
it is the diameter across the roots of the threads, p is the pitch : 
this is the distance the bolt will advance in the direction of its axis 
for one complete revolution. The pitch is also the reciprocal of 
the number of threads per inch. is the thread angle. 

35. Standard Screws. — No extensive explanation is needed to 
point out the necessity for complete interchangeability of screws of 
the same nominal dimensions. 

To secure this interchangeability all countries have adopted stand- 
ard systems of screws, bolts and nuts. Unfortunately, however, the 
systems of different countries are not like each other. Thus, the 
United States standard screw has an angle of 60°, while the British 
has an angle of 55°. There are also other differences, and for some 
of the nominal sizes the number of threads per inch, or the pitch, is 
different. 

In such a system of standard screws a limited number of diam- 
eters is selected, and for each diameter a fixed pitch is determined, 
while the thread must have an accurately defined form. 'Eygyj 
screw must very accurately conform to the standard nominal and 
effective diameter and pitch, and the thread angle is also important. 
However, for many sizes the United States nut will go on the British 
bolt, when the pitch is the same, but the fit between the two is not 
accurate. The width and depth of the head and nut is also fixed by 
the nominal diameter of the bolt, thus permitting the use of stand- 
ard wrenches. 

The table in the back of the book gives the standard system used 
in the navy, and generally throughout the United States. 

In the absence of the table, or for sizes not tabulated, the pitch 
and effective diameter may be found from the following formulae: 



p = 0.24Vd + 0. 625-0.175, approximately. (24) 

di=:0.91d-0.08, approximately. (25) 



6 



78 



ENGINEERING MECHANICS. 










•iP/, 







^ 







Fig. 37. 



Fig. 38. 



J3i///re^^ r/^r^67c/ 



/f/zuc/r/e T^rea€/ 




Fig. 39. 



Fig. 40. 



/^//>e 77;reac/ 



^60y 




Fig. 41. 






BOLTS. Is^UTS. SCREWS. 79 

36. Forms and Proportions of Screw Threads. — Figs. 36-41 show 
the forms and proportions of threads commonly used in practice. 
Fig. 36 is the United States standard, or Sellers Y thread, used for 
bolts and nuts. The angle of the thread is 60°. The pitch is deter- 
mined by the formula given in Art. 35, but in practice is always 
obtained from a table. One-eighth of the depth is cut off from the 
top and bottom, forming a flat point, which tends to prevent burr- 
ing of the thread. The tap for cutting the nuts has a diameter 
slightly greater than the screw, to allow a small clearance. 

Fig. 37 is the square thread. The pitch for standard square- 
thread screws being approximately twice the pitch of the standard 
V screws of the same nominal diameter. The pitch is given by the 
formula 

p = l/n = 0.16d-f 0.08, nearly. (26) 

With this form of thread there is no bursting pressure on the nut, 
as the thrust is parallel to the axis of the screw. These threads cost 
more to cut than triangular threads, and cannot well be formed by 
dies. They are very generally used for transmitting power. The 
depth of the thread of the screw is sometimes made \^ p, in which 
case ^p is the depth of the thread in the nut, thus giving a slight 
clearance. 

Fig. 38 is the acme thread, which is a modified form of square 
thread, the angle of the thread being 29°. The slight taper thus 
provided greatly facilitates its being rapidly engaged and disengaged 
when used with a split nut, as in the screw-cutting lathe. 

Fig. 40 is the round or hnucMe thread. This makes a very strong 
screw, and one that will stand much rough usage without damage. 
It is not suitable for transmitting power. 

Fig. 39 is the trapezoidal or buttress thread. This form combines 
the important advantage of the square thread, that of having its 
thrust parallel to the axis of the screw, with the strength of the V 
thread. The power, however, can only be transmitted efiiciently in 
one direction. It is an excellent form where there is little or no 
work to be done by it in a reversed direction; as, for example, the 
screw of a jack, in some form of presses, and for the breech block of 
large guns. 

Fig. 41 is the standard pipe thread, known as the Brigg^s system 
of pipe threads, used in the United States. The following is a 



80 EXGIXEERIXG MECHANICS. 

synopsis of the standard table, giving the numbers of threads per 
inch for the various sizes : 

I" pipe has 27 threads per inch 

i" and f" pipe has 18 

r and f" pipe has 14 

r to 2" pipe has lU 

2V pipe and over has. . . 8 '' cc u 

Tlie pipe sizes refer to the nominal inside diameters of the pipe. 

37. A Feiu Practical Applications of Bolts, Studs, Xiits, etc. — 
Fig. 42 is the ordinary stud, used most commonly in securing the 
cylinder head to the flange of the cylinder, and other similar con- 
nections. The distinction between a bolt and a stud is that the latter 
is threaded at each end, and one end is permanently screwed into 
one of the pieces to be connected, and a nut is then screwed on the 
other end. As indicated in the figure, the width of the flange should 
be three times the diameter of the stud, and the thiclniess one and 
one-half times. 

Fig. 43 is a forcing holt, used to break the joint under a cylinder 
cover, or a follower ring, before lifting with eye bolts. 

Fig. 44 illustrates a case where a stud with a nnt on each end 
would have to be used, instead of the ordinary headed bolt. 

Fig. 45 illustrates a case in which both a stud and a tap holt must 
be used. 

If a stud is used at A it will be impossible to use one at B also, as 
the angle plate C could not be put on or taken off. Consequent!}', a 
screw must be used at B which is put in after the plate is put over 
the stud at A. 

Fig. 46 shows a flanged nut, which is sometimes used when the 
bolt hole is considerably larger than the bolt. This is an expensive 
nut to manufacture, and in many cases the same purpose may be 
served by using a washer. 

Fig. 47 shows a flanged cap nut, used to prevent leakage through 
the screw threads. 

In places where a nut is subject to vibration, as on many parts of 
moving machinery, it tends to slack back, or even work off its bolt 
altogether, however perfect its fit might have been originally. One of 
the most common methods of preventing this is to fit a second nut, 
screwed down on the first, as shown in Figs. 48-50, to jamb or lock 
it. Fig. 48 shows the proper method of fitting a lock nut. A little 



BOLTS. NUTS. SCREWS. 



81 



Ordinary Stud. 



Forcing Bolt. 





Fig. 42. 



Fig. 43. 



Stud Nutted at Each End. 



Stud and Tap Bolt. 




Fig. 44. 



Fig. 45. 



82 



ENGINEERING MECHANICS. 



consideration will show that the top nut may take the whole of the 
load, and, therefore, this is the proper place for the thick nut. The 
faulty arrangement shown in Fig. 49 is often seen in practice, hut 



Flanged Nut. 



Cap Nut. 




Fig. 46. 



Fig. 47. 



it is none the less faulty. This arrangement has come ahout prob- 
ably from the fact that most wrenches are too thick to set up the 
thin nut when it is at the bottom. The obvious way of escaping this 
difficulty would be to make both nuts full thickness, but there may 
not always be room to do this, and it is rather unsightly in appear- 
ance. This has led to the compromise shown in Fig. 48, where both 



Good Practice. 



LOCK NUTS. 
Faulty Practice. Convenient Compromise. 







LqA 




I — -z^ 
Fig. 50. 



I of the thickness of a 



nuts are of equal thickness, each from f to 

full-size nut. In all cases, wherever it can possibly be done, the 

method of Fig. 48 should be used. 

In cases where the jarring is apt to be excessive, the upper nut is 
still further secured by driving a pin through a hole in the end of 



BOLTS. NUTS. SCREWS. 83 

the bolt, the upper surface of the nut having a slot into which the 
pin fits. If, after several readjustments, the nut has to be screwed 
further down on the bolt, a washer is fitted under the pin, of a thick- 
ness just enough to require the pin to be forced in. 

Other methods of locking are by means of a set screw through the 
side of the nut ; the use of a locking plate ; the use of a collar nut, 
with a locking ring, or set screw, similar to the method familiar to 
all, shown on Sheet 4 of the course in mechanical drawing. 

38. Comparison of Square and V Threads. — If the tensile load 
on a Y-threaded screw acting parallel to its axis be represented by 




Fig. 52. — Forces Acting on a Screw Thread. 

W (Fig. 51), then E will represent the pressure acting normally to 
the surface of the thread (this is also a measure of the friction), 
and H will represent the force tending to burst the nut. It is evi- 
dent, from the triangle of forces, that the larger the angle the 
greater is the bursting pressure, and the larger is the amount of 
friction. When the angle decreases and the surface of the thread 
becomes perpendicular to the axis of the bolt, the bursting pressure 
disappears, and the normal force E becomes equal to W. This is 
the case of the square-threaded screw, and, therefore, the square 
thread is used to transmit power. On the other hand, for the same 
depth of thread in each case, the V thread has about twice the 
amount of material resisting shearing at the root of the thread, ab, 
as the square thread. Therefore, the Y thread is stronger, and is 
generally used for fastenings, where strength is the primary con- 
sideration. It will now be apparent why the buttress thread com- 
bines to a considerable extent the special advantages of both the Y 
and the square threads. 



84 ENGIl^'EERING MECHANICS. 



BOLTS. XUTS. SCREWS. 



85 



39. Fatigue of Bolts. — Where bolts are subjected to repeated 
shocks, or the action of live loads, it is foiind that they frequently 
fail by breaking across the threads, even when carrying a less load 
than they have safely carried before. The cause of such failure is 
attributed to a slight temporary elongation at the weakest part, the 
threads, each time the stress occurs. This risk of this failure may 
be reduced to a minimum by making the bolt of uniform strength 
throughout its length (a) by turning down the body of the bolt to 
the diameter of the bottom of the threads, or (b) by drilling a hole 
up from the head through the axis of the bolt to the point at which 
the threads begin. In either case the object is to make the effective 
area of the bolt the same throughout its length. In case method (b) 
is used, the size of the hole is obtained as follows : 

Let d = nominal diameter of bolt, or diameter of unthreaded part. 



6-1 = effective diameter of 

threads. 
d' = diameter of hole. 



bolt, or diameter across roots of 



Then 



rd'2 _ nd,' 



or 



d'z=Vd2-di^ 



40. Straining Action on Bolts. — In Fig. 
52 a bolt is represented as carrying a load 
W, without initial tension in the bolt, be- 
fore the load is applied. Let f be the safe 
working stress allowed for the material of 
the bolt, and let d^^the effective diameter 
of the bolt. Then 

W= ^' f, 



or 



d,= 



\/4W 

rrf 



(27) 



(26) 



Load with no Initial 
Tension. 



from which the size of the bolt can be 
determined. 

In Fig. 53 let C and D represent two 
rigid flanges which when bolted together 
form a fluid-tight joint. In order that the joint may remain 




86 ENGINEERING MECHANICS. 

tight after the load W is applied, the flanges must not separate 
to the slightest extent. This condition can exist only if the 
initial tension produced in the bolt by screwing np is equal to 
or greater than W.* Thus it is seen that the strength required in a 
bolt is not directly dependent on the load it is to carry, but is also 
dependent on the strain produced in making the joint. 

Fig. 54 represents diagrammatically the case of a joint made with 
elastic packing, the elasticity of the packing being represented, for 
clearness of description, by the springs. Then, if the initial thrust 



Load with Initial Tension. 
Elastic Joint. 



Load with Initial Tension. 
Rigid Joint. 





of each spring is represented by -JTi the initial tension in the bolt 
will be Ti, due to screwing up the joint. 

If we call the extension of the bolt a, and the compression of the 
springs b per unit of load, the initial compression of each spring 
is ^T^xb, and the corresponding extension of the bolt is T-^xa. 

If a weight W is applied, the tension in the bolt is increased to a 
value T between W and W + T^, and the additional tension of the 
bolt diminishes the compression of the springs. If W is increased 
until the additional extension of the bolt equals or exceeds the 
initial compression of the springs, the thrust of the springs dis- 



* This statement will be more fully explained a little later. 



BOLTS. NUTS. SCREWS. 87 

appears, and the tension of the bolt is that due to the load alone, 
as represented by Fig. 52. In making a fluid-tight joint, therefore, 
the bolts must be screwed up tight enough initially to prevent the 
thrust of the packing being exceeded when the pressure is acting. 

When the load is added, the additional extension of the bolt 
becomes a(T — T^) and the compression of each spring diminishes 
toi[bT,-a(T-T,)]. 

If we call E the thrust due to hoth springs in this condition, we 
have 

bE=bT,-a(T-TJ, 

E = T,--^(T-TJ; 
but in this condition T = W + E, whence 

Avhich reduces to 

bT = Wb + T^b - aT + aT^, 

(b + a)T = Wb+(b + a)T„ 

Tr=-^W-i-T,. (28) 

b+a ^ ^ ^ 

If the springs are easily compressible (corresponding to a joint 
made with very soft and elastic packing) so that b is large com- 
pared to a, the term in equation (28) approaches unity, and 

b -[- a 

the tension, when the load is applied, becomes W -f T-l approximately. 
If the compressibility of the springs and the extensibilit}^ of the 
bolts are equal (a case that may be approximated when a very hard 
gasket is used) b becomes equal to a and the tension in the bolt 
then becomes ^W + T^. If the joint is very rigid (as when the joint 
is made by grinding the flanges to true surfaces, and no gasket is 

used), b becomes small compared to a and the term is a very 

small fraction, and the tension is not appreciably increased by the 
application of the load, and is then equal to T-^ approximately. 

41. Friction and Ejficiency of Square-Threaded Screws (modified 
from TJnwin). — Fig 55 represents a square-threaded screw con- 
necting two parts, A and B, whose approach to each other is resisted, 
either because B carries a load W, or because A and B are the 



88 



ENGINEERIN-G MECHANICS. 



flanges of a joints the elasticity of the packing between them resist- 
ing compression. This resistance is symmetrically disposed, and 
its resultant is an axial force T producing tension in the body of the 
screw. Suppose that the screw is tightened by the application of a 
force F acting at an arm 1 — for instance, by a wrench on the bolt 
head. 

Let d be the nominal diameter and d^ the effective diameter of 
the screw and D the diameter of the collar^ or head. Let ix = tan<f> 




Fig. 55. — Friction of Screw. 



be the coefficient of friction between the threads and nnt, B; and 
fji=taB. </)', the coefficient of friction between the collar, or head, and 
A. If r is the mean frictional radius of the thread, and E the mean 
frictional radius of the collar, we have : 

"^-^2 + 2 
and, similarly, 

R=i(d + D). 



= i(d + dj; 



BOLTS. NUTS. SCREWS. 89 

Let p be the pitch of the screw, and a the angle of inclination of 

the thread to the normal, at the radius r. Then tana= -7——. 

27rr 

If there were no friction the pressure on the thread would act 
normall}^ to it, or along the direction ac (Fig. 55), which makes an 
angle a with the axis of the screw. In consequence of the friction, 
however, the pressure is deflected to an angle <!> (such that tan <^ = />i) 
to the normal to the thread, or it acts along the direction ad, whose 
angle with the axis is a + c^. We now have the screw in equilibrium 
under the action of an axial force T along ab; a horizontal force 
H at radius r along bd due to the force applied to the spanner ; and 
a pressure Q between the threads of the bolt and nut, along ad, 
making an angle a + (/)-with the axis. Then, from the triangle of 
forces, we have 

T:H:Q::ab:bd:ad; 
or T/H = ab/bd; but ab/bd = cot(a + (^), 

.•.H = Ttan(a + c/>). (29) 

The moment of H about the axis is Tlxr (r being the arm at 
which the horizontal component of the friction acts). Therefore, 
we have : 

Moment, M=:Hr = Tr tan(a + <^). (30) 

From trigonometry tan(a + d))= ana+ an (/> ^^^ 
° J V ^/ 1 -tan a tan 

= p/27rr, and tan c/> = /.; from (30), M == Hr = Tr ( }^^^ ^ + ^^'^ ^ 
^' ^ r. \ /> Vl-tanatan</> 

= Tr I ''-' ' ' I = Tr ( g + ^'^^ V (31) 

2nr 

The friction of the collar is /T acting at a radius E, its moment 
is, therefore, 

H'E = /TE. 

The force F applied to the wrench, acting at a leverage 1 balances 
the frictional resistances of the screw. Thus we have 

Fl = Hr + H'E. (32) 



90 ENGINEERING MECHANICS. 

E-fficiency of Screw. — The useful work done in tightening the 
screw, or driving against a resistance T is Tp per revolution (that 
is the force T acting through the distance p) and the work expended 
on the wrench is F x SttI. Therefore, the eflSciency 

-p _ useful work ^ Tp ,r.n\ 

total work F27rl * ^ ^ 

The value of Fl may be obtained from equation (32). 

Note. — ^When the screw is used to lift a weight, as in the screw 
jack, the expression T in equations (29), (30) and (31) becomes 
W, the weight to be raised. 

If the friction of the collar is eliminated, which may be accom- 
plished in the case of the screw jack by allowing the weight to 
revolve with the screw, the expression for eflQ.ciency becomes 

E= ^ -£ . Since now Fl = Hr (from equation 32), dividing nu- 
merator and denominator by 27rr, we get (from equation 29) 

■p _ 27rr _ T tan a _ tana ,r,A\ 

TF ~ T tan(a+<^) ~ tan(a+'^) ^ ^ 

Maximum Efficiency. ^Fiom equation (34) it is seen that E 
becomes zero when a = and when a = 90 — <j> and must, therefore, 
have a maximum value between these limits. 

This maximum efficiency occurs when a = 45°— -L , and we then 

z 

have 

tan(45<'->-) tan (45° -4) 

Maximum B= -^ "^ ^ = ) =-\- (35) 

tan (45» - ^ + <l>) tan (46° + -| ) 

The proof of this is long ; and the easiest way of arriving at it is to 
solve equation (34) with several values of a, plot a curve and pick 
out from it the maximum value. 

In case the screw is fitted with a standard size nut, or has a head 
of standard size (the size being determined by the nominal diameter 
of the screw) and the thi-ust of screwing up is taken by the nut, or 
the head, we have this friction to consider in addition to that of the 
threads alone. In the case of the nut, or the head, the friction acts 
at a radius of approximately IJ times that of the threads. 



BOLTS. XUTS. SCREWS. 



91 



Then H'E = /.'TE=:f/x' Tr( since E = 14 Xr) ; if ix= fx' =taii<f> this 
becomes 

H'R=fTrtan</>, 

and equation 32 reduces to 

Fl = Tr[tan(a + </>)+f tan0]. 
For Y threads. 



\27rr — piu,sec)8/ 
(see Fig. 56). 

j^rc. ■ ,_ tan a(l— //, tan a sec^) 
" ~ tan a + ixsec /3 
Comparing these with similar equations for square threads, it is at 
once evident that considering efficiency and mechanical power, 
square threads should be used in preference to V threads. 



(36) 
(37) 
(38) 
(39) 




Fig. 56. 

42. The Coefficient of Friction of Screws. — Experiments have 
been made to determine the friction between square-threaded screws 
and nuts of different metals and with different lubricants with the 
result that no marked differences in the friction were found with 
different metals or different pressures. The friction varied with 
different lubricants, as shown in the table. 



TABLE 12.— COEFFICIENT OF FRICTION. 


Coefficient ft = 


Min. 


Max. 


Mean. 


Lard oil 09 


.25 
.19 
.15 


11 


Heavy mineral machinery oil ... . .11 


.U 

.07 







43. Tensile Stress Produced in Bolts in flaking Joints. — If the 
resistance in screwing up a joint is that due only to the spring of the 



92 



ENGINEERING MECHANICS. 



packing which, is being compressed, the total tension T in the bolt 
depends only on the moment El exerted by the wrench. A series of 
direct experiments have been made at Sibley College, by which it 
was found that skilled mechanics would use an effective wrench 
length of very nearly 15 times the nominal diameter of the bolt, and 
would graduate the pull on the wrench in proportion to the diameter 
of the bolt. The average measured tension in the bolt was about 

T = 16,000d, (40) 

although in some individual cases the tension was much greater 
than this. Using equation (40) as a basis, the initial stresses pro- 
duced in bolts of various sizes in making steam-tight joints are as 
follows : 



TABLE 13.— STRESSES IN BOLTS IN MAKING JOINTS. 



Nominal diam. of bolt, d 

Total tension, T 

Stress per sq. in., of effective area, lbs., f . 



%" 


1'' 


2" 


10,000 


16,000 


32,000 


49,500 


29,100 


14,000 



4" 

64,000 

6430 



In the case of the f-inch bolt it will be seen that the breaking 
strength for iron, or the softer varieties of steel, is practically 
reached and for the 1-inch bolt the stress is dangerously near the 
yield point of such mei^ls. These experiments are not accepted as 
absolutely correct, but they show that, for the smaller sizes of bolts, 
the initial tension produced in making the joint and before any 
steam pressure is acting^ may easily overstrain the bolts. For this 
reason it is customary in making such joints to use nothing smaller 
than say f-inch bolts. 

The above-mentioned experiments were conducted by sJcilled 
workmen. Such a man, it was found, would instinctively select a 
wrench of a length about 15 times the diameter of the bolt. On 
board ship it is frequently necessary to have less skillful men make 
joints, and experience shows that such men have a tendency to use 
too long a wrench and to exert too great a pull for the size of the 
bolt. This is a further practical consideration fixing a minimum 
size of bolt to be used. 

In order to provide for the above stresses, and also to allow a mar- 
gin of safety to provide for shocks such as those produced by water 
in steam cylinders, as well as the additional stress due to torsion 
of the bolt, a relatively small working stress is allowed in calculating 



BOLTS. NUTS. SCREWS. 



93 



the sizes of such bolts. In other words, a large factor of safety is 
used. 

In practical designing the engineer has available some standard 
reference, or pocket book, containing tables to guide him in selecting 
the safe working stress allowable. In the absence of such a complete 
table the following may be used : 

TABLE 14.— (SPOONER.) SAFE WORKING STRESS FOR BOLTS 

AND STUDS. 

Finished Joints. 



Values of f. 



Steel. 



Iron. 



Bolts and studs 1" nominal diam. and above ' 6,000 

%" " " " under (4,500 to 3,000) 

" " " ordinary marine practice 5,000 

" " " cylinders less than 10" diam 2,500 



4,800 

(3,600 to 2,400) 

4,000 

3,000 



For joints with rough ilan^ges and with comparatively thick gas- 
kets, the values given in the above table (Table 14) may be reduced 
one-half, to insure safety. 

44. Additional Stress due to Torsion. — It is shown in Art. 41 
that the friction of the threads of a screw produces a moment about 
the axis of the screw, whose arm is r (the mean radius between the 
outside and the root of the thread) and whose force is H; or a 
moment Hr. Now this kind of moment is called a twisting or 
torsional moment and produces internal stresses in the body of the 
screw in a manner somewhat similar to that of a beam subject to a 
bending moment ; the stress produced being in this case, however, in 
the nature of shearing. As in the case of a beam we have the gen- 
eral formula M = f Z, so now, in the case of a twisting action, we have 

M = f.Zp, (41) 

in which Zp may be called the polar modulus of section, being the 
polar moment of inertia divided by the distance of the most distant 
fiber from the axis. 

In the case of the screw the twisting moment produced in its 
body is the moment applied by the wrench minus the moment of 
friction of the head (or nut). That is, the friction of the head 
7 



94 ENGINEERING MECHANICS. 

absorbs a part of the total effort, leaving only the remainder effective 
in producing twisting in the body of the screw. Then M=:Hr 
(see eq. 30) ; the polar moment of inertia of a solid circular section 
is Ip=7rDV32 and Zp = VD/2=7^DV16 = 0.196D^ In this case 
D is the effective diameter of the screw, or d^ of Fig. 53. Substi- 
tuting the above values in (41) we have 

fs = M/Zp=Hr/0.196di3. (42) 

The usual proportions of square threads are such that di = 0.8d 
(approximately), hence 

' rr=i(d + d,) = -^J^=.45d (42) 



so becomes 



but 



. _ Hx.45d _ 4.4H ... 
*^- 0.196 x.512d« - ~W- ^^^^^^-y)' 



H = Ttan(a+<^) 
see eq. (30) ; thus 



^ _ 4.4Ttan(a + (^) ,^gv 



The total tension is T and we have, then, 



whence 



therefore, 



4 < X '1 



ft= 5 (nearly), (44) 



4.4Ttan(a + </)) 
^' ^' —=2.2 tan (a + </»). (45) 



2T 

d^ 



An average value of a may be taken as about 3°, and a value of /* 
of .12, corresponding to a value of <^ of about 7°, or a + cj> = lQ° ; and 
tan(a + <^) =0.176, substituting this value in (45), gives 

^ = ,387 or say 0.4 (nearly) . (46) 

Thus we see that in tightening up a bolt or a screw, there is a 
shearing or twisting stress produced in addition to the direct ten- 
sion, of a value of about y^ of the direct tension. 



BOLTS. 2^UTS. SCREWS. 95 

These two stresses must be combined to find the resulting stress 
in the bolt. Now the maximum principal stress is found from the 
equation 

fma. = if. + 4V4f? + f?" (47) 

(from Smith, " Strength of Material/^ with the symbols changed to 
accord with the method used in this book) . Substituting in equa- 
tion (47) the value of fs = 0.4f^, we get 

f^a,,=:1.14f,. . (48) 

In other words the resultant stress due to the combined tension 
and twisting increases the stress hy IJ^fo for the ordinary square- 
threaded screw. In the case of the standard V thread the increase 
of the stress is about 16 fo to 17^. 

45. Steam^Tight Joints. — In making steam-tight joints, such as 
the joint under the cylinder head or valve chest cover, the studs or 
bolts must be made strong enough to carry the load due to the steam 
pressure acting on the area of the cover and also the initial tension 
produced in setting up the nuts in making the joint. They must 
also be spaced near enough together to prevent leakage under the 
flange between them, owing to the flange being sprung or opened 
by the action of the steam pressure. When studs are used the width 
of the cylinder flange should be 3d, where d is the nominal diameter 
of the stud. When bolts are used the width of the flange should be 
greater than this. The pitch to be used (that is, the distance from 
center to center of stud or bolt), depends on the steam pressure 
used, and also to some extent on the thickness of the flanges. The 
following gives a good rough approximation for the pitch : 

For H. P. cylinders, pitch=3.5d. 
LP. " " =4.5d. 

L.P. " " =5.5d. 



In general, pitch = J 



loot 

P 



where P = pressure in pounds per square inch. 

t=i thickness of cover in sixteenths of an inch. 

The thickness of the flanges are at least equal to the diameter of 
the stud or dolt, and if possible they should be from IJ to IJ times 
d. As pointed out in Art. 43, the use of a bolt or stud of less diam- 
eter than f inch is to be avoided, so it is evident that most joints of 



96 ENGINEERING MECHANICS. 

steam pipes and small cylinders will have a large excess of strength, 
as far as the steam pressure alone is concerned. 

Example. — Calculate the size, number and pitch of the steel studs 
for a cylinder cover : diameter of cylinder, 24 inches; steam pressure, 
180 pounds per square inch. 

From the pressure this is evidently a H. P. cylinder, and, therefore, 
we may pitch the studs about SJd. Then the diameter of the pitch 
circle will be about 24" + 3d, since the least width of flange should 
be 3d. As a first approximation, assume a stud diameter of 1 inch, 
then the diameter of the pitch circle of the studs is 27 inches and 

number of studs=— ^--^ =24.2, say 24. (It is customary to use 

an even number of studs for symmetry.) 

It is usual to assume that the steam pressure acts on a circle 
whose diameter is the pitch circle; that is, it is assumed that the 
steam pressure extends somewhat beyond the inner edge of the 
cylinder flange. N'ow, equating the total strength of the studs to 
the total load on the cover, we have 

24X^^ Xf=-^^X180. 

From Art. 43 (Table 14) we find that a safe value for f is 5000 

27^X180 
pounds per square inch. Therefore d^^ = — — whence d^ 

= 1.045. This is the effective diameter. 

Eef erring to the table in the back of the book, we find the cor- 
responding nominal diameter to be between l-J and 1 J inches, there- 
fore, we will use 24 IJ-inch studs. The width of the flange will 
then be 3| inches and the diameter of the pitch circle 27f inches 

and the actual pitch= ""^^^/^^ =3.567 (about). 

Then 

Ans. d = lj"; ^^ = 24; p = 3.567 (about). 

Questions and Problems. 

Make a neat sketch of the U. S. standard screw thread, showing 
the nominal diameter, the effective diameter, the pitch of the thread, 
the angle of the thread, the amount of flattening at the point and 
root of the thread. Explain the importance of a system of standard 
screw threads (Arts. 34, 35, 36). 



BOLTS. NUTS. SCREWS. 97 

Sketch the various forms of screw threads commonly used, and 
explain in general the relative nses and advantages of the several 
forms. Discuss the use of lock nuts (Arts. 36, 37). 

Explain clearl}^ why the square-threaded screw is better to use 
for transmitting power than the V thread. What is the fatigue of 
bolts due to ? How may the risk of failure in such cases be mini- 
mized (Arts. 38, 39). 

Discuss the straining action of bolts due to setting up on a joint 
with and without elastic packing. Illustrate by supposing the 
flanges separated by springs. Deduce an expression for the thrust 
in the bolts in terms of the initial thrust, the load, the compression 
of the elastic material and the extension of the bolt (Art. 40). 

A square-threaded bolt is tightened by a force F applied with a 
leverage 1, producing an axial force T in the bolt. Calling r and R 
the mean frictional radii of the thread and head respectively, fx the 
coefficient of friction, p the pitch of the screw and a the angle of 
inclination of the thread; deduce an expression for the moment Fl 
in terms of,T, r, p, jx and a. Let E = ljr. Also deduce an expres- 
sion for the efficiency (Art. 41). 

Discuss the stresses produced in bolts in making joints and show, 
in a general way, why it is inadvisable to use a bolt less than about 
I inch diameter (Art. 42). 

Deduce an approximate relation between the shearing stress and 
the tensile stress in a square-threaded screw. Make the following 
assumptions: Ip = 7rDY33; di = 0.8d; total direct thrust=:T; 
a=3°; ii = .12 (Art. 44). 

Problems. 

1. A man in setting up on a square-threaded tap bolt 1 inch 
diameter exerts a force of 40 pounds on the end of a wrench 15 
inches long. Assuming that the friction of the bolt head acts at a 
radius E = 1.4r; diameter across roots of thread = 0.68 inch; pitch 
= 0.24 inch; coefficient of friction = 0.125; find the total thrust pro- 
duced in the body of the bolt and the efficiency. 

2. A man in setting up on a square-threaded tap bolt 2 inches diam- 
eter, exerts a force of 80 pounds on the end of a wrench 30 inches 
long. Diameter at bottom of thread = 1.63 inches; pitch = 0.40 
inch; coefficient of friction = 0.12 5. Assuming that the friction of 
the bolt head acts at a radius E= 1.4r, find the total thrust produced 
in the body of the bolt, and the efficiency. 



98 ENGINEERING MECHANICS. 

3. Two men in setting up on a square-threaded tap bolt 4 inches 
diameter exert each a force of 80 pounds on the end of a wrench 
60 inches long. Diameter at bottom of thread = 3.33 inches; pitch 
= .72 inch; coefficient of friction =0.125. Assuming that the fric- 
tion on the bolt head acts at a radius E = 1.4r, find the total thrust 
produced in the body of the bolt^ and the efficiency. 

4. A man in setting up on a square-threaded tap bolt 2 inches 
diameter exerts a force of 80 pounds on the end of a wrench 30 
inches long. Diameter of bolt at bottom of thread=1.63 inches; 
pitch of thread = 0.40 inch; coefficient of friction = 0.125. Assume 
that the friction of the bolt head acts at a radius of E = l.]-r 
[r=J(d + di)]. Find the tensional and shearing stresses produced 
in the bolt. From these find the maximum stress produced. 

5. The diameter of a steam cylinder is 30 inches ; steam pressure 
= 180 pounds per gauge. Find the number, size and pitch of the 
studs for the cylinder cover (steel studs) . 

6. The diameter of a steam cylinder is 32 inches ; steam pressure 
= 200 pounds per gauge. Find the number, size and pitch of the 
studs (steel) for the cylinder cover. 

Note. — In this case use a stud diameter of 1-J inches for first 
approximation. 

7. The diameter of a steam cylinder is 74 inches; working steam 
pressure = 50 pounds per gauge. Find the number, size and pitch 
of the studs (steel) for the cylinder cover. 



CHAPTEE YII. 



The Screw Jack. 

46. Practical Prohlem 11.^ — Design a screw jack to raise a load 
of fifteen tons. Height of lift 12 inches. The frame is to be of 
cast iron ; the screw, head, nut and band of wrought iron. The head 
of the screw is to have two IJ-inch 
holes at right angles to each other 
for the turning bar. There is to be 
a cast-iron swivel plate on top of 
the head. The screw must not 
overhaul. 

1. To Find the Diameter of the 
Screiv. — This is calculated for the 
bottom of the thread (as the threads 
do not assist in resisting the com- 
pression due to the load). 
Let A = area of cross section of 
screw ; 
W = total loader 15x2240 = 

33,600 pounds; 
M = PI = turning moment re- 
quired to raise W. 
This moment will induce a torsional 

stress ia= ^^ = -^, which must 

be combined with the pure com- 

W 

pressive stress f c = -^— , and this 

combined stress for wrought iron 

must not exceed 10,000 pounds per 

square inch (assuming that the 

screw is frequently loaded and unloaded, and not subject to shock or 

reversal of stress). 




Fig. 57. 



* From " Notes on Machine Design." 



100 EXGINEEKING MECHANICS. 

In the preliminary calculations, we will assume that an increase of 
20^^ in the area (over that required for pure compression alone) will 

be a sufficient allowance for the effect of the torsional stress fa,* 

qo p,r\r\ "I o 
or A= ' X — ^ — =4.032 square inches, and d^ (effective diam- 
eter) = 2:27, or 2-3V", and A becomes 4.08 square inches. (Mg. 57 
shows a half -sectional view of the screw jack with all parts as- 
sembled.) 

2. To Find the Dimensions of the Threads. — For a screw which 
will not overhaul, the pitch angle must be less than the angle of 
friction, i. e., tana<tan<^. 

Threads will be made square for the best transmission of power 
and for durability. To be safe against overhauling with the mate- 
rials used and good lubrication, fx must not be given a greater value 
than 0.10. 

.-. c^==tan-i ^ =tan-i yV =5°45', and a<cf>. 

From t3.no = -^^, substituting proper values (see ISTote 1) and 

solving we get p = 0.84, which is the maximum value to satisfy the 
conditions that the screw will not overhaul. 

The value of p thus found is a check only, to insure that the screw 
will not overhaul. Practically, with a pitch so great, the pull on 
the turning bar would be excessive ; to reduce the pull on the bar the 
pitch is reduced and its value is found as follows : 

From equation 25, Art. 35, di = 0.91d — 0.08 (approximately). 
By substituting the value of d^ in this equation we get 2^ = .9 Id 
— .08 or d, the nominal diameter = 2 14 inches. This value of d is 
only approximate, but is close enough to be used in the calculation 
of the pitch. From Art. 36 

p = .16d-{-0.08 
= . 16 X2-Lf 4-0.08 = ^496 or -1". 

To Find the Nominal Diameter. — For square threads, consider 
the depth of the threads -J pitch, then 

d=d,-hp.-.d=2 3V'+r=2ir.t 

* See Art. 44. 

f For buttress threads, the angle of the thread is 45° and the depth 
of the thread is |p, after ip has been taken off at point and root of 
the thread (see Fig. 39). That is to say d^di + fp. Be careful to 
make the flat part of buttress threads the bearing surface. 



THE SCREW JACK. 101 

3. To Check the Calculations for Cross Section of Screw. — The 
moment required to raise load W is equal to M=:Pl = Hr = Wr 
/ p + 27rr/i \ ^g^g ^^^^^ ^^^^ ^^.^-^ ^^^^^g 

33,600 xliifi^t^^^^l^^ii^) =6970 inch -pounds. 

\ ^TI" X J- ^4 2" X TO" / 

This is the moment which produces the torsional stress 

M , ^ Trdi^ . , 6970x16 „„„_ 

ia = ^; and Z=-j^.-.ia = ^^^^^2^^ =3000 

pounds per square inch, and this torsional stress must be combined 

with the pure compressive stress, 

, W 33,600 Q^^^ T . -, 

fc= — r— = ' =8230 pounds per square inch, 
A 4. (Jo 

in order to obtain the combined stress f on the screw which is given 

by the formula 



*f = .35fc + .65Vfc' + 4:fa' = .35x8230 + .65V 8230 + 4x3000 ' 
= 9500 pounds. 
As this is well within the limit of safet}', the area of cross section of 
screw previously calculated will be taken. 

4. To Find the Dimensions of the Nut. — The height of the nut 

is determined from the equation W = Kxnx ^ X (d^ — d^^), in 

which K is the maximum allowable pressure in pounds per square 
inch on bearing surface, and its value depends on the speed. 

For a rubbing velocity of less than 50 feet per minute (and it is 
fair to assume that the screw will never exceed this limit) , the value 
for wrought iron is K = 2500. Then the number of threads required 
in the nut will be 

W _ 33,600 



, = 6.76. 



TT 



Kx ^ (d^ - d,^) 2500 x ^ [(^H)^" (2A)T 

Height of nut = pxn = ^x6.76 = 3.38", or 3f" (a short calcula- 
tion will show that the direct shearing stress on the threads will 
come well within the limit allowed for wrought iron, with ample 
allowance for wear). 

The nut is screwed into the inside of the frame with a pipe thread, 
which need not be calculated, as the nut is screwed hard up against 

* This is a modified form of equation (47). The constants .35 and 
.65 are derived from a consideration of Poisson's ratio, and their deduc- 
tion will be given in a later chapter. 



102 



ENGINEERING MECHANICS. 



the shoulders a and b and against the band at c (Fig. 59). B is a 
wrought-iron band shrunk on the frame to prevent the nut and 
frame from spreading. 

The dimensions are empirical, and must be decided upon for each 
size of jack. For this size of jack the band is J inch thick and 1^ 
inches long. The outside diameter of band = the outside diameter 
of the nut = the outside diameter of frame + -J the thickness of band. 

5. To Find the Dimensions of the Read of the Screw. — The area 
remaining after the holes are taken out is made equal to the area 




Fig. 59. 



of the screw at bottom of thread, as it must sustain about the same 
stress. 

Fig. 58 is a horizontal section taken at the middle of the holes. 
Let x = diameter required. 

A = area of screw, at bottom of thread, 
d = diameter of hole in the head ; then 

=rA+[2(x-d)d + d2]=:A+(2dx-d-) since two lengths 



nxr 



d across, less the central portion dxd, are removed. 
In this case, substituting proper values. 



^x2=:4.08+(ljx2x-lixli). 



Solving, we find 



x = 4r04, or 4^''. 



THE SCREW JACK. 



103 



The depth of the head below the holes is calculated on the assump- 
tion that the end of the screw tends to punch upward in to the head, 
and that the resistance to this is one of pure shear (which is on the 
safe side, as there is an additional resistance to crushing offered by 
the small centrally projecting portions in the vicinity of the holes). 

Let ty Fig. 60, = thickness required such that 

7rd,tf, = W, 
where is is the maximum shearing stress for wrought iron = 7500. 
Solving, we find 

,_ W _ 33,600 



rdjfs 



X 2^X7500 



,625, or 



It is customary to make the distance from the top of the hole to 
the top of tlie head the same as that below = f inch, 

. • . total depth of head = 2 X f + 1 J = 24". 

6. Details of the Swivel Plate. — The central pin is a part of the 
screw head, and for this size of jack is taken as l-J inches diameter. 
It stops just below the level of the top of the swivel plate, the latter 
being countersunk. It is arranged to turn by dropping it into place 
and then slightly upsetting the central pin over the top of the plate. 
Height of swivel plate = 1 inch; upper diameter = 5 inches; and 
lower diameter = 4rJ inches. Depth of counter bore, y\- inch; diam- 
eter of counter bore, 2 inches. These dimensions Yary according to 
the size of jack. By calculation, 

the bearing surface may be worked 
out and compared with the crush- 
ing strength of cast iron. It is 
ample in this case. 

7. Details of the Frame. — The 
diameter of the inside of the 
frame is that of the outside of 
the screw plus -| inch to yV i^^^^ 
clearance on each side. In this 
case it is 2|f''-f i" = 3i". (Area 
corresponding =7.2 square in- 
ches). The outside diameter is 
found as follows : 

Consider the crushing strength 
of the cast iron of the frame equal 
to that of the material of the screw. Then, for equal strength : 




Fig. 60. 



104 EN"GIiSTEEKING MECHAi^ICS. 

Area of screw at bottom of thread = area of outside diameter of 
frame minus area of inside diameter ; or 

Area of outside diameter of frames area of inside diameter 
+ area of screw. 

Or 7^2 + 4^08 = 11.28 square inches, and 

Outside diameters 3': 79, or Sff". 

Thiclaiess of frame = -J (outside diameter — inside diameter) 

— 1 2'f 

This is for strength only, and if used it would be necessary to 
increase the thiclaiess at the top of the frame for the threaded por- 
tion of the nut; also, in order to avoid the danger of breaking 
accidentally from rough usage, and to insure a better casting the 
thickness is increased to f inch. 

The outer diameter of frame is then = 3-3^^" + l^" = 4-3y. 

The diameter of the base of the frame is made about 2^ times the 
above or lOff for a good spread to prevent tipping, and for a larger 
surface. The base is made thicker so as to press on a greater quan- 
tity of material. It is finished in some such manner as shown in 
sketch. The bottom of the frame must be at least -J inch below the 
bottom of the screw when at its lowest position in order to protect the 
screw. 

The inside diameter of the frame at the upper end is counter- 
bored J inch all round for a length of 2 inches and threaded for the 
nut, terminating in a square shoulder against which the bottom of 
the nut is screwed up hard. Also the outside diameter at the top of 
the frame is turned down -J inch all round (over a length of 1-J 
inches) to a square shoulder on which the band rests. 

Note 1. — From tan a= — ^2 — we have p = 27rr tan a. 
Eemembering that for square threads, depth of thread tt; P ~ 9 

nearly, and that r^ — radius at bottom of thread, and r = r^ + -^ ^ 

P 
4 



'^ttU 



4 



H- ^-l,tana 



and substituting 



_ 27rri tan a 



1 — ^ tana 



106 



ENGINEERING MECHANICS. 



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THE SCREW JACK. 107 

7. Check the height of the nut to prevent stripping of threads. 

8. Find the dimensions of the head. 

9. Calculate the inside and outside diameters of the frame, and 
its thickness. 

10. Lay down the jack on the drawing board, working out the 
minor details, such as the swivel plate, the reinforcing band, etc., as 
the work progresses. 

Note. — Instead of making a complete half section it is suggested 
that a "break" be made in the frame a short distance below the 
bottom of the nut, and only the part above the break shown in half 
section. Continue the outline of the screw below the break by 
broken lines, so as to be able to record its length, which is an im- 
portant dimension, on the drawing. The object of this is to avoid 
the tedious work of actually drawing the threads over the whole 
length of the screw, and to reduce the amount of cross hatching. 
The legend reads : 

Screw Jack. 

Capacity, tons. 

Designed by (name), Midshipman, First Class. 
Date. 



CHAPTEE YIII. 

Compound Stresses. 



BENDING WITH TENSION OR COMPRESSION. COMBINED STRESSES. 
STRUTS AND COLUMNS. EULER^S FORMULA. EMPIRICAL FOR- 
MULiE, RANKTNE^S OR GORDON'S. STRAIGHT LINE FORMULA. 

49. Bending Combined tvith Tension or Compression. — Fig. 61 
represents the stresses produced in a bar under a load P acting 

normally to the cross section AB and 
parallel to the axis of the bar, but to 
one side of the axis and at a distance 
X from it. We may introduce two 
equal and opposite forces P' and P", 
each equal to the original force P, 
acting against each other along the 
axis, without disturbing the condi- 
tions. We thus have the bar subject 
to the direct compressive force P" 
yf^'^ and a couple whose moment = Pxx. 

Let A be the area of the cross section 
of the bar. We then have a stress 

P 




Fig. 61. 
Action of Eccentric Load. 



distributed over the section : 



due 



to the direct thrust and a stress 

M Px 

± — - = d= --y=— at the edges due to the couple. See equation (18), 

Art. 24. Adding these, the total stresses on the edges of the bar are 

P Px 
A "^ Z • 



f: 



(49) 



The greater stress is of the same kind as P ; the other may be the 
same, or opposite, according to the value of x. 

The distribution of the stress is shown by the shaded part 
ABB' A' of Fig. 61, A A' and BB' representing the stresses along the 
edges. 



COMPOUND STKESSES. 109 

If the bar is of a circular section^ diameter d, we have by sub- 
stituting in equation (49) 

f = P/-^ ±Px/-32-= ^(1± ^^ ] (50) 

If of rectangular section, side li in plane of bending, and b at 
right angles, equation (49), reduces to 

. P _^ Px P /. _^ 6x\ ..-,., 

From equation (50) it is seen that the stress along one side be- 

8x 
comes zero and along the other side is doubled when - , =: 1 or when 

d 

Similarly for the rectangular section the stress along one side is 

zero and is doubled alonp- the other when . - =1 or x= -^ . 

ho 

It will be negative, or of opposite sign to P along one edge when 
the value of x is smaller than the above ratios. 

In attempting to find the dimensions of a section to sustain a 
given eccentric load P, wdth an assumed allowable value for the safe 
working stress f, from equation (49) it will be found that a trouble- 
some cubic equation results. In such cases the practical way is to 
assume a trial section and check this for P and f. In the case of a 
solid circular or square section, where but one dimension is unknown, 
equation (49) can be solved. 

Example. — A crane has a swing of 30 inches, and has the cross 
section shown by the figure (Fig. 62). The safe working stress of 
the material (for compression) is 9000 pounds per square inch. 
Find the load W the crane can safely lift. 

The section being symmetrical, the neutral axis will be at a dis- 
tance of 2 inches from the edge of either flange. The moment of 
inertia I is 

3-V[2x(4)^-1.5x(3)^]=7.3. 
.*. Z rr I /y = 7.3/2=1 3.65. A==2x4-1.5x3 = 3.5. 



No^ 



f = -^ -f — ^ , equation (49)^ 



110 



ENGINEERIN-G MECHANICS. 



whence 



W = 



fAZ 

Z + Ax ' 



^^ 9000x3.5x3.65 ^ 9000x3.5x3.65 ^-^qqq j|^g 



3.65 + 3.5x30 



108.65x1 




Small Crane. 



This method is used also for the design of large hooks, the frames 
of machines, such as punching machines, etc. 

50. Combination of Stresses in, General. — Let the elementary 
particle shown in Fig. 63 have a parallelopipedal form, the length 
of the top and bottom sides being B and the ends xA, and the thick- 



COMPOUND STRESSES. ill 



112 



ENGINEERING MECHANICS. 



ness being uiiity. Assume the ends A to be subject to a unit tensile 
(or compressive) stress f* and at the same time to the unit shearing 
stress fg. This shearing stress tends to rotate the particle, but the 
tendency to rotate produces a shearing stress along the sides, which 
balances that along the ends. The area of each side is B x unity ==:B. 
Let fs' be the unit shearing stress produced along the sides. Then, 
since equilibrium is sustained, the moment of the forces along the 
ends balances that of the forces along the sides; therefore, fsA (the 
total force at each end) xB (the arm of the couple) = the mo- 
ment, or 

f,AxB = f;BxA.'.f, = f;, 



Y^^ ^^/I 




A^^ 



Sj;,3=j^^^ 



Fig. 63. — Compound Stress. 



or a shearing stress at the ends involves a shearing stress of equal 
intensity at the sides of the particle. Let C be a diagonal of the 
particle, making the angle <^ with the top and bottom sides. This 
diagonal section is subject to an aggregate shearing force qxC 
(q being the intensity of shearing stress along C), and to an aggre- 
gate tensile force pxC (p being the intensity of the tensile stress). 
Resolving all the forces acting into directions along and parallel 
to C, we have 

q X C = f i A cos <^ + f sB cos <^ — f sA sin <^, ( 52 ) 

p X C = f i A sin (^ -f f sB sin (/> + f s A cos <^ ; ( 53 ) 



COMPOUND STRESSES. 113 

from which we get 



but 



ABA 

q = f^ ^ coscf> + is ^ C0& <f>-ts ^ sine/), 



A/C = sin ct>, and -^- = cos (f> ; 



hence 

q = ft sin <f> cos cf>-{- f ., cos^ </> — f s sin^ <^, ( 54) 

p = f / sin- (/> + f s sin (^ cos <^ + f « sin ^ cos </>. ( 55 ) 

These equations reduce to 

q = -Jff sin 2(/) + fs cos 2<^, (56) 

p=-ifj(l-cos 2</)) -j-ts sin 2</>. (57) 

Solving for maxima 

d(2.#.) 
from which 



— 0=^it cos 2</) — is sin 2</), 



tan 2(/)= Y ±01' ina^- q^ (58) 

similarly^ from (57) 

^-^ ^Ozzzif, sin 2</> + f, cos 2<j{.; 
whence 

tan 26=- ^ for max. p. (59) 



When 



_ ¥t 



tan2cf>=^ 
sin 2(f> = 



ii 



and 

Substituting these values in (56) we have 




which reduces to 

q.,«.= Vif7+"f?. (60) 



114 ENGINEERING MECHANICS. 

Similarly, by substituting the values from (59) in (57), we get 



*Pma..=if^ + VW+f?, (61) 

*p,„,„z.:if,-V|f7+f7. (62) 

51. Struts or Columns. — A short compression member, whose 
ratio of length to. lateral dimensions is not greater than 4 or 5 will 
fail, when overloaded, by direct crushing, if loaded centrally. The 
effect of eccentric loading of such short struts is discussed in Art. 
49. When the length of the column, in proportion to its lateral 
dimensions, becomes great, failure will occur by lateral bending, 
unless the column is absolutely straight, with the load acting exactly 
along the axis of the center of figure of the section, and the material 
is absolutely homogeneous. These conditions are impossible to 
obtain. Therefore, the nature of the stresses are those of a bar sub- 
ject to both bending and compression at the same time. 

The first approximately satisfactory rules for the resistance of 
long thin columns were obtained by Euler, and these are still used 
where the ratio of length to least lateral dimension is about 30. The 
formulas obtained by Euler are based on the elastic theory of mate- 

* In tlie screw-jack problem a formula for combined stress was 
given in the form 



f = .35ff + .65V4f/ + ffMArt. 46, 3.). 

This form of the formula is obtained as follows: 
From equations (61) and (62) the principal stresses are 

Pi = ift + iV4tV+ty 
and 

p, = -^f, — ^V4f/4-tV. 

The deformation produced by these stresses is ^ — a^ where X= — 

Ji. Hi m 

= Poisson's ratio. 

The equivalent normal stress, that is, the single linear stress that 
would produce the same deformation as that actually produced by the 
combined stress is 



Pe = p, — Xp, = 4fi + ^V4f/ + ft== — |Xf( + i\V4f«»+f*^ 
= ^f^ (1 — X) +i(l + X)V4f7+f?. 
When X = .25 this reduces to 

l?e=%tt + %V4f7+^'. (63) 

When X = .30 we get 

Pg — .35fj-f.65V4f?^l?. (64) 



area; 7r^ = 10. 


Then 




W = 7r^^ =10 -f- (nearly 


and 






W lOIE lOEk^ 
P~ A AP P 



COMPOUXD STRESSES. 115 

rials, and their deduction is given in Smith's " Strength of 
Material." 

Eulers Formula.'^ — Let E = the coefficient of elasticity of the 
material; A = area of section of the column; I = the moment of 
inertia for bending; k = the least radius of gyration of the section 
such that P^I/A; L = the actual length of the column; l = the 
length of arc of the curved bar measured between two points of 
contrary flexure; W = the greatest load consistent with stability; 
p = W/A = the corresponding greatest stress per unit of sectional 



(65) 



(66) 

Equations (65) and {Q^) are Euler's general equations. The 
ratio of 1 to L has different values, depending on the way in which 
the ends of the strut or column are held. The following tabular 
diagram shows the usual cases, together with the corresponding 
modifications of formulae (65) and (66). 

The feet of the columns in Fig. 64 are shown bolted down to 
emphasize the importance of having the ends rigidly fixed. If the 
feet are flat flanges merely resting on the supporting surfaces, the 
columns are little better than those with rounded ends. 

In the formulae given in connection with Fig. 64, W is the ulti- 
mate load the column can carry without failure, and the greatest 
safe working load is obtained by dividing W by a factor of safety 

W 

N". The safe working load is, then, ^^ , and the safe working stress 

f must not exceed -^ . The value of f must not exceed the value 

of the safe crushing resistance of the material fc. 

The commonly used values of the factor of safety are: 

Cast Iron Wrought Iron. Timhpr 

i^astiron. Mild Steel. limbei. 

N=::8. 5s^=5. N=10. 

When there is doubt that the load acts in the axis of the column, 
the above values of X should be increased by one-half. 

* For the deduction of Euler's formula, see Smith, " Strength of 
Material." 



116 



enghsteering mechanics. 



DIAGRAM OF STRUT LOADING AND EULER'S FORMULiE. 



Case 1. 

One end fixed, other 
end free. 



l = 2L 




Case 2. 

Both ends rounded 
or pivoted. 



Case 3. 

One end rounded or 

pivoted, other 

end fixed. 



2.5EI 
2.5Ek2 




W//////////^ 



W 



lOEI 
L2 



10Ek2 



W 



20EJ 



?0Ek2 



L2 



Case 4. 
Both ends fixed. 



fc^ 







w = 



40EI 
L2 



40Ek2 
L2 



Fig. 64. 



52. Limit of the Application of LJulers Formuloe. — The principal 
defect in Euler's formulae is that they neglect the direct compressive 
stress in the material, and give only the load that v^ill cause failure 
by bending. Even this load is given only approximately, and the 
load calculated by them is too great, because the bending moment 
increases more rapidly than the formulas assume, due to the fact that 
after bending once starts, the load no longer acts in the neutral axis, 
because the neutral axis shifts when bending starts (see Art. 49). 
However, in very long columns in which the compressive stress is 
small compared to the stress due to bending, the error is not serious. 
Therefore, they are used for such very long columns when the 
virtual length 1 is not less than 30 times the least lateral dimension 



COMPOUND STRESSEiS. 117 

for wrought iron or steel : or is not less than 12 times for cast iron 
or wood. 

If applied to ver}^ short, or moderately short, columns or stmts, 
whose ratio of length to lateral dimension is below about 10, they 
give results for the load which are absurdly too high. Thus, for 
example, take an iron strut, 4 inches diameter and 40 inches long, 
with rounded ends. Then 
lOEk^ 



p^ 10X29,000,000X1 ^ ,81,000 pounds per square inch. 

This result is absurd, as the value is far higher than the crushing 
strength of the material. 

53. Empirical Formuloe for Struts or Columns of Moderate 
Length. — When a strut is shorty that is, when the ratio of length to 
least lateral dimension is about 5 or less, its strength, or rather the 
load it can support, is obtained by the relation, Wc=:Axfc, where 
Wc = the load; A = the area of cross section; and fc = direct crushing 
resistance of the material. When the column is long, that is, when 
the ratio of length to least lateral dimension is about 30 or more, 
Euler's formulae give satisfactory working approximations. When 
the ratio of length to lateral dimension is between the limits of 5 
and 30 approximately, neither Euler's nor the crushing formula 
give sufficiently accurate results. Various empirical formula have 
been proposed to cover such cases. Of these the best known is one 
devised by Gordon and made more general by Eankine. It is an 
interpolation formula, which gives results agreeing with the crush- 
ing formula when the strut is very short, and with Euler^s formula 
when it is very long, and at the same time gives results agreeing 
with actual experiments for lengths between very short and very 
long. 

54. Rankine's Formula. — For a short strut, where buckling is 
impossible, we have 

Wc = Axfc. (67) 

For a long column, with the ends free, Euler's formula is 

W6 = 7r^EI/L2=:7r2EA(k/L)^ (68) 



118 



ENGIXEERIXG MECHAIs^ICS. 



Now, if Wc is the ultimate load for any length L and cross-section 
A, we may assume an equation 



1 



1 + 1 



If the strut is very short this equation will evidently give a value 



all 



of Wa, which will approximate to Wc, because ^ff^ becomes so sm 

as to be negligible, since a small value of L in equation (68) will 
make Ws large. Consequently, Wa = Wc very nearly. Again, if the 

strut is long, ==^^ becomes negligibly small in comparison with ^™r- 
Wc Wb 

and Wa^Wft very nearly. Now, since the change in Wa caused by 

changing L is continuous, for a fixed value of A, it is reasonable to 

assume that equation (69) will give a proper value of Wo for any 

length of strut. 

Then, for a strut with free ends we have 

1 1 



Wa = 



+ 



1 



We ' Wb 

fc-A 
f T 2 
1 + 



TT^Ek^ 



fc.A ' rr'El 

fc.A 

1 + a 



(70) 



where a = 



TT^E 



and is, therefore, a constant for any given material. 



The symbols have the same meaning as given in Art. 51. 

TABLE 15.— VALUES OF fc AND a FOR USE WITH RANKINE'S 

FORMULA. 



Material. 



Cast iron 

Wrought irou 

Mild steel 

Hard steel — 
Timber 



fc=lbs. 

per 
square 

inch. 



80,000 
36,000 
55,000 
70.000 
7,200 



Case 1. 
One end 

fixed, 

other end 

free. 



_1_ 

400 

1_ 

2250 

T3V5 



Case 



Both ends 
pivoted. 



1600 

1 

9000 

5T00 

o"'0 0'0' 

YoO 



Case 3. 
One end 

Hxed, 
other end 
pivoted. 



18000 

1 

11000 

1__ 

10000 

_ 1 

1500 



Case 4. 



Both ends 
fixed. 



6100 

1 

36000 



2 007 

__1 

3000 



The value of the constant varies for the different methods of end 
fixing; thus for a column fixed at both ends (Case 4 of Fig. 64), 



COMPOUXD STRESSES. 11!) 

the value is a/4; one end fixed and the other end free it is 4a (Case 1 
of Fig. 64) ; one end fixed, the other end pivoted it is a/2 (Case 3 
of Fig. 64). 

Table 15 gives the values of fc and a commonly used. 

I^OTE. — The values of a in this table are to be used as given, in 
equation (70), since the differences due to the different kinds of 
end supporting have been applied to the constant. Also in equation 
(70) L is the actual length of the column from center to center. 

Equation (70) gives the breaking load for a column, and to 
obtain the breaking or buckling stress, W is divided by A, the area 
of cross section. Thus we have for the breaking stress p 

The safe vjorJcing stress is, then, p/N, where N is the factor of 
safety. In the case of connecting rods, the load being a " vibrating '' 
one, the value of IST is between 8 and 10. 

The practical application of the above formulae is more fully 
shown in the solution of the connecting rod problem. 

55. Straight Line Strut Formula. — Experiments on full-size 
struts and columns show that, unless the most elaborate precautions 
are taken to insure dead accuracy of loading, and unless the material 
is perfectly homogeneous, the results may easily be from 10^ to 15^ 
out when compared with the results as calculated from such for- 
mulae as Eankine's, or Gordon^s, or similar ones. Even when every 
possible precaution is taken the results are often 5fo or more out. 
For these reasons many engineers prefer the use of a straight line 
formula of the form 

p = M-N-i-, (72) 

where p = buckling load in pounds per square inch. 

M = a constant depending on the strength of the material. 
Nrzia constant depending on the form of section and the 
elasticity of the material. 
l = the equivalent length of the strut. 
d = the least lateral dimension of the strut. 
The following table gives the values of M and N. 



120 ENGINEERING MECHANICS. 

TABLE 16.— VALUES OF M AND N FOR STRAIGHT-LINE FORMULA. 



Material. 



Form of section. -s- not to exceed 




Mild steel. 



Hard steel. 



30 



M. 



47,000 
47,000 
47,000 
47,000 



71,000 
71,000 
73,000 
71,000 



114,000 
114,000 
114,000 
114,000 



N. 



soft cast iron. 



Hard cast iron. 



Timber, 



90,000 


4,100 


90,000 


4,700 


90,000 


3,900 


90,000 


5,000 


140,000 


6,600 


140,000 


7,000 


140,000 


6,100 


140,000 


8,000 


8,000 


470 


8,000 


500 



COMPOUND STRESSES. 



12.1 



The straight line formula is very commonly used for columns 
and in bridge designing, etc. For piston rods and connecting rods, 
however, the Bureau of Steam Engineering uses Eankine's for- 
mulae, or a modification of them, as will be given under the practical 
connecting rod problems. 

Questions and Problems. 

Discuss the action of a compressive load when its line of action is 
parallel to, but not in, the axis of the load. Deduce equations for 




Fig. 65. 



the stresses produced (1) for a circular section; (2) for a rectangu- 
lar section. 

Show how to find the load a simple crane, such as a boat davit, can 
safely carry. 

A piece of material is under the action of a tensile and at the 
same time a shearing force. Show how to find the resulting maxi- 
mum shearing and tensile stresses produced. 

Given Euler's formula p=: ^ , explain the meaning of the 



122 ENGINEERING MECHANICS. 

symbols. Make sketches showing the four usual cases of end sup- 
porting and show how the above formula is modified in each case. 

What is the serious defect of Euler^s formula ? Show how Gror- 
don's or Eankine^s formula is deduced given Euler's formula 
,.. lOEI 

Problems. 

1. A boat weighing 8 tons is supported by two davits as shown in 
the sketch. f = 6000. Find the diameter of the vertical part of the 
davit. 

2. A hollow cast-iron column, fixed at each end^ is 20 feet high, 
and has a mean diameter of 12 inches. It is to carry a load of 100 
tons. Factor of safety 8. What is the thickness of the metal? Use 
Eankine's formula. 

3. A solid mild steel column, with round ends, is 6 inches in 
diameter and 37 feet long. What load will it bear? Solve by 
Euler^s and by Eankine^s formulae and compare the results. 

4. A tie bar of rectangular section, 2 inches wide and 1 inch thick, 
carries a tensile load of 10 tons, which acts at a distance of yo i^^h 
from the axis of the bar, in the direction of the thickness, but is in 
the center of the width. Find the extreme stresses. 

5. A short cast-iron strut, 8 inches external diameter with the 
metal 1 inch thick, carries a load of 20 tons which acts If inches 
from the axis of the strut. Find the extreme stresses. At what 
distance from the axis must the load act to cause tension along one 
side of the strut. 

6. Find the principal stresses and their directions in a beam at a 
point where the unit normal stress is 400 pounds per square inch 
and the unit shearing stress is 250 pounds per square inch. 

7. A round steel propeller shaft is subject to a normal stress, due 
to bending, of 5000 pounds per square inch, and to a shearing stress, 
due to torsion, of 8000 pounds per square inch. Poisson^s ratio = .25. 
Find the. equivalent stress. 



CHAPTEE IX. 

Cotters. 

56. Cottered Joint. — Fig. 66 repre- 
sents the cottered joint very often used 
to connect two rods A and B which are 
required to transmit a force of tension 
or compression in the direction of their 
length. The end of rod B is made in the 
form of a socket into which is fitted the 
end of rod A and the two are held firmly 
together by driving in the cotter C. 
The clearance at G, H and I is a very 
important point to be considered in 
designing such a joint as is shown in 
Fig. 66. It is obvious that since the 
cotter is to draw the rod A into the 
socket there must be clearance at H 
and I, and also an equal amount at G. 
The usual amount allowed is from yV 
inch to I inch. 

The taper of the cotter is also im- 
portant. It will be shown later that 
the total taper should not exceed 1 in 
7, but in practice, especially if no pro- 
vision is made for locking the cotter, 
the taper is usually much less than this ; 
about from J inch to ^ inch to the foot 
being common practice. 

The material of the cotter is prac- 
tically always steel, whatever may be 
the material of the rods. 

57. Method of Proportioning the 
Parts of a Cottered Joint for Uniform 
Strength. 

Let f^ = safe working strength in tension of the rods, pounds per 
square inch. 




Pig. 66. 



-Cottered Joint. 



124 



ENGINEERING MECHANICS. 



fs=:safe working strength of the rods in shear, pounds per 
square inch (fs = from 0.7f# to 0.8ft for steel, or 0.5ft 
for iron). 

fc = safe working crushing strength of the bearing surfaces^ 
pounds per square inch. 



f/J /foe/ Teors. f^2J f^/ie/o/Zfoe/ ^€/rs. (3) SocA^e^ 7hc7rs. 

// 



f^Co^/er S/eetirs. 





(SJ^yzc/q/^/^oe/Cri/s/ges 



ri^ 7/. 






/Y^./7^. 



//J^/pe/^^oc/re/ S/ze(7rs. ^'9^^- 



^" 



fi^.Z£. 



W^ 



^a^ 



1 



%>zj: 



&)f:Ai(/o//foc/6^/ iecrrs. /S!J Oo/Zar C'ri/s/ies. (^OJ Collar SAee/rs O// 
"^ // 



V 



T 





y^^ 7^. 



fY^. 7S. 



/^i^.76'. 



Illustrating Failure of Cotter Joints. 



fg'=:safe working shearing stress of the cotter, pounds per 

square inch. 
P = safe working load of the joint. 
The above figures, 67 to 76, illustrate the various ways in which 
the several parts of the joint may fail either in tension or compres- 
sion. When the load acts alternately in compression and tension, 
the safe working stresses are only one-half those for either tensioii 
or compression alone. 



COTTERS. 1^5 

In a properly designed joint of this kind the resistance to failure 
in each of the above illustrated ways should be equal, so that every 
part has the same strength. The diameter of the main body of 
the rods A and B is first found and the proportions of the parts of 
the joints are found in terms of d. 

Resistance to Tension. — (1) The rods may fail by tearing at A 
or B (Fig. 67) . We then have 

V^-^it, (73) 

or 



d 



/4P 



(74) 



(2) Failure may occur by the end of the rod tearing across the 
cotter hole (Fig. 68). Then 

P=(4'-d,t)f,; (75) 

equating (73) and (75) we hai^e 



rcl^ - ^d,^ dt 
4 4 



But it is usual to make t^^d^, 



whence 



• • 4 4 4 ^ 

7rd2=(7r-l)di^ 



. 1 / TT _ . /3.1416 

di = 1.21d. (76) 

To find the value of t in terms of d we have 

t=ii- = 14^ = 0.3d. (77) 

4 4 ^ 

(3) The socket may tear across the cotter hole (Fig. 69). Then 

p^[^_!L(5!z:d^_(D_eljt]ft. (78) 
Equating to equation (73) 

'^f, = [|(D^-d,=)-(D-d,)t]f,. (79) 



126 ElS^GINEERIXG MECHANICS. 

Substituting the values of d^ and t 

.785d2 = .785(D2-1.464d2) - (D-1.21d) x.3d. 

Simplifying and transposing, 

785D2 - 300Dd - ISTld^ = 0. 
D'^-.382Dd-2d- = 0. 

Completing the square, 

D2-.382Dd + .0365d2 = 2.0365d^ 
D-.191d = 1.428d, 

Drrl.62d. (80) 

For additional stiffness and for convenience it is usual to increase 
this somewhat, say 

D = 1.75d. (81J. 

(4) The cotter may fail by double shearing (Fig. 70). 
The mean breadth of the cotter is b. Then 

P = 2btf/; 
equating to equation (73), 

2btf/=:-^ft, 
2bx.3dxf/ = .785d2f^, 

b = 1.31d-/4-. (83) 

Is 

When all parts of the joint are of either wrought iron or steel, 
the breadth of the cotter is usually 1.6d. 

(5) The bearing surface of the cotter in the rod end may fail by 
crushing (Fig. 71). 

P::=ditf,. (84) 

This equation is to be used as a check on the values of d^ and t 
already found. If the value of P obtained from equation (84) 
works out less than the load for which the joint is being designed, 

then d-L and t must be increased, keeping the proportion, t= —^ . 

(6) The bearing surfaces of the cotter in the socket may fail 
by crushing (Fig. 72). 

P=(D,-di)tfe. (85) 



COTTERS. 127 

Equating this to the crushing of the rod end, 

(D,-dOtfo = d,tfc, 
D,-d, = d,, 

Di = 2di = 2 X 1.21dr= 2.42d. (86) 

(7) The cotter may shear through the end of the socket (Fig. 73). 

P = 2[l(D,-d,)f«], (87) 

= 2[l(2.42d-1.21d)fs]=2.421dfs; 

equating to equation (73), 

2.42klf,= ^d2f,; 

the shearing strength of wrought iron parallel to the fibers is about 
.bit; then 

l = .65d. (88) 

If the rods are of steel f., may be taken equal to .8f ^ when 

l:=.43d. (89) 

In practice it is usual to increase these values to : 

for wrought iron, l=:.75d. (90) 

for steel, l = .5d. (91) 

(8) The cotter may shear through the end of the rod (Fig. 74). 

P = 2lAfs, (93) 

= 2.42lidf«, 

which is exactly the same as the case above. So that 

li = l. (93) 

When the joint is in compression (9) it may fail by crushing the 
collar E (Figs. 66 and 75). 

P=^(d/-d,^)f.. (94) 

In order to keep equal strength, the bearing surface on the collar 
must equal that of the cotter on the end of the rod, so 

^(d,= -d,^) =ii_J- since t=id„ 

3.1416d/ = -J:.1416di^ 

d2=:1.15di = 1.15xl.21d = 1.4d. (95) 



128 



ENGINEERING MECHANICS, 



But in practice this is usTially increased to 

d2 = 1.5d. 
(10) The collar may be sheared oi? the rod (Fig. 76). 

P = Trdit-^f s ; 
equating to equation (73), 



(96) 
(97) 



substituting 



but di = 1.21d and ts = OMt, 
ti = .413d, say 0.42d. 



(98) 




"I 




Fig. 77.— Effect of Cotter Without Gib. 




Fig. 78. — Gib and Cotter. 



58. Gib and Cotter. — ^When a strap DE is to be held on a rod G a 
plain cotter, such as shown in Fig. Q6 cannot be used, but a gih and 
cotter (sometimes called gib and key), as shown by Fig. 78, must be 



COTTERS. 



129 



fitted. If the cotter without a gib were used, the friction between 
the strap and cotter at the point C, Fig. 77, 
would cause the former to be sprung away as 
shown by the broken lines at B. The use of the 
gib H, Fig. 78, prevents this action and is the 
method in common use for connecting the strap 
to the end of the connecting rod for many slow- 
moving engines. 

It should be noted that when the gib and 
cotter are used the holes in the strap and rod are 
parallel, the taper being given to the back of the 
gib, as shown at AB, Fig. 78. The dimensions 
shown in Fig. 78 refer to the total breadth b of 
the gib and key, and are the usual proportions 
given the parts, having been found satisfactory 
in practice. In order to prevent the key slacking 
back while at work, it must be locked fast in 
some manner. Fig. 79 shows one method of 
doing this, the screw shown being rigidly at- 
tached to the gib. Another simple method is to pio. 79, Method of 

fit a set screw through the body of the rod bear- Locking the Key. 
ing directly on the side of the key. 




130 



EXGINEEEING MECHANICS. 



59. The Taper of the Cotter. 

Let 11= total horizontal force necessary to drive the cotter home 
when the connected parts are under a load P. 

a = angle between the edges of the cotter, in other words, 
the taper of the cotter in degrees. 

(^ wangle, of repose of the materials, or the friction angle. 



It is then evident from Fig. 80 that 

H = H, + H, = P[tan(c/) + a)+tan<^]. 



(99) 




Fig. 80. — Force to Drive 
Cotter Home. 



Fig. 81. — Force to Drive 
Cotter Out. 



This is the force necessary to tighten up the cotter under the 
load. 

Similarly from Fig. 81, the force necessary to loosen the cotter is 

H = H, + H, = P[tan(c^-a)+tanc/>], (100) 

but the cotter will just slip back without any additional force when 
H = 0: then 



COTTERS. 131 



or 



P[tan((/)-a) +tan </>] =0, 

tan(<^ — a) = —tan 4>, 



tan cb — tan a , , 

~~ , = —tan d), 

l + tan</)tana 

tan <^ — tan a= — tan cj) — tan^ </> tan a, 
2 tan<^ = tana(l— tan^ cf)) ^ 



tana=.-^^5i_ z=tan2<^, 
1 — tan^ci) 



or 



a = 2</>. (101) 

Thus it is seen that when the angle of the taper of the cotter is 
equal to twice the friction angle the cotter will be just on the point 
of working loose ; and to prevent this occurring a must be less than 
2cf>. 

For slightly greasy metal the value of cf> is taken at 4J°, therefore, 
a must be less than 9°^, which corresponds to a taper of 1 in 7. In 
practice, as already stated in Art. 56, the taper is made from | inch 
to ^ inch to the foot for safety. This corresponds to a taper of from 
1 in 48 to 1 in 24. 

Questions and Problems. 

Make a neat sketch of a socket and cotter joint for connecting 
two round rods. Explain the various ways in which the joint may 
fail under the action of a tensile or compressive load. 

Show how to find the proportions of a socket and cotter joint in 
terms of d the diameter of the rods, to resist both tension and 
compression. 

Make a neat sketch showing the use of a gib and key. Explain 
why it is necessary to use a gib. In a gib and key connection, show 
a method of locking the key. 

Find the greatest taper that can be given a cotter so that it will 
just not slide out of place when the load is acting ; given the friction 
angle for slightly greasy metal as 4-J°. 



132 engineering mechanics. 

Problems. 

1. Two rods are to be connected together by a socket and cotter 
joint. The load is a steady pull of 10 tons. Design the joint, all 
parts to be of mild steel. Use Table 8 for the values of the working 
stresses. 

2. A long pump rod is made of two parts connected by a socket 
and cotter joint. The load is an alternate tensile and compressive 
one of 5 tons in each direction. The rod is so guided that buckling 
may be neglected. Material, rods, phosphor bronze; cotter, mild 
steel. Use working stresses given in Table 8. Design the joint. 



CHAPTEE X. 

The Theoey of the Connecting Eod.* 

60. The connecting rod of a steam engine is subjected to rapidly 
alternating, and nearly equals tensions and compressions. Theo- 
retically it may fail in one of three ways : 

1st. As a "long column'' in compression, by bending out the 
line of its axis^ i. e., " buckling." 

2d. As a " tie rod " by pulling apart under tension. 

3d. As a " short column '' in compression, by the metal break- 
ing down or " crushing.'' 

Practically we need consider the rod only as a long column in 
compression, because the formula for strength against failure by 
buckling gives a larger section than for either of the other tM^o 
cases. 

61. Gordon s Formula. — The formula for long columns^ known 
as Gordon's formula, has been given in a great variety of forms. 
One of the most general forms for a column " free at both ends " 
has been used in the Bureau of Steam Engineering for the tabula- 
tion of the designs of the connecting rods of all the U. S. naval 
vessels. 

It can also be used for designing rods, or special formulas de- 
duced from it for special cases. 
It is 

- Ix^ 

W= '^ ^. (102) 

1+— X-^ 

An intelligent use of it requires an understanding of the terms 
and the values given to them under different circumstances. 

A, the area, and k, the radius of gyration, of the cross section 
of the rod are the unknown factors whose values are to be found 
by the application of the formula for one or more places along the 
length of the rod. 

* From " Notes on Machine Design." 



134 



ENGINEERING MECHANICS. 



The Bureau of Steam Engineering uses the following method to 
determine the load W the rod is designed to bear: It is supposed 
that on starting the engine, before the receiver spaces have been 
filled with steam, the piston of the H. P. cylinder may be subjected 
for one or two strokes to full boiler pressure on the upper surface, 
while the lower surface is under atmospheric pressure only. E, the 
thrust on the H. P. piston rod, is then the area of the H. P. piston 
X boiler pressure. W, the thrust on the connecting rod, is greater 
than E by a small percentage. The difference is a maximum when 
the crank is at right angles to the axis of the cylinder and the 
connecting rod in consequence makes its greatest angle with that 



axis. This angle is of course sin~^ 



crank length 



connecting rod length 



(see Fig. 



24'' 
82) and with the proportions there given is sin"^ — ^ = sin"^ i 

= 14° 29'. At the moment considered, the cross-head pin is acted 




Fig. 82. 



5 Wcose~T 



Fig. 83. 



on by three forces, W, the thrust of the connecting rod; E, the 
thrust of the piston rod; and S, the resistance of the guides (see 
Fig. 83). Neglecting friction, i. e., taking S perpendicular to E 
and resolving forces horizontally, we have Wcos^ = E, or W = E 
sec B. 

In marine practice the crank is very frequently one-fourth the 
length of the connecting rod. In all such cases 

W = E sec 14° 29' = Ex 1.033. 

This calculation applies only to the H. P. cylinder, but in nearly 
all cases the rods are made equal and interchangeable. 

C 

The factor ^r^ is often written as one svmbol, f , and is defined as 

the " safe working fiber stress." However, since C also appears 
separately, tlie older method of using C, the " ultimate strength " 
of the material, and providing a " factor of safety " N is adhered 
to as the best in this case. 



THE THEORY OF THE CON]\TECTiNG KOI). 135 

The ultimate strength of the metal used is largely a matter of 
cost, and constantly changes with improved methods of manufac- 
ture. For slow-running or small-sized engines, where saving in 
weight is not of prime importance, wrought iron, on account of its 
cheapness is still used. For quick-running engines the tendency 
has been towards the best attainable material. This tendency is 
well shown by the specifications of our naval vessels here briefly 
tabulated. 

Name of material. Ult. Str. Vessels. 

Wrought iron 36,000 Puritan and Miantonomoh. 

Mild steel 55,000 Boston to Dolphin. 

Steel 65,000 Maine (old) to Iowa. 

Oil-tempered steel 80,000 Kearsarge to Illinois and destroy- 
ers. 

Oil-tempered nickel steel.. 95,000 Maine (new) and all later vessels. 

The first steel used was of very low carbon, 0.20,^, for the sake 
of great ductility and consequent safety against fracture from 
sudden shock. 

The process of oil-tempering has enabled steel of 0.30 to 0.40;^ 
carbon to be used, with its attendant great strength and without 
loss in ductility. It is requisite, however, that the thickness of the 
metal tempered should nowwhere exceed 2J inches to three inches, 
and, consequently, /or large rods oil tempering necessitates a hollow 
section. 

The presence of a small percentage of nickel in the steel, makes 
it possible to reach an ultimate strength of 95,000 pounds. Without 
it, it is barely possible to get 80,000 pounds. 

The factor of safety must be a large one by reason of the '^ yi- 
brating load " on the rod. Its value usually lies between 8 and 10. 
Eecently the Bureau of Steam Engineering has used values of 8.5 
and 8.9. 

TV is, of course, 3.1416. tt-, in G-ordon^s formula, is, however, 
taken as 10 for simplicity. 

E is the modulus of elasticity of the metal of the rod. For all 
iron and steel rods its value is taken as 30,000,000 pounds. 

1, the length of the column when the formula is used to de- 
termine the section at the middle of the rod is the length '' from 
center to center." 

62. The Rod Considered as a Column. — The strength of a coluam 
to resist buckling is known to depend largely on the nature of the 



136 ENGi:f^EERIN"G MECHANICS. 

" end fastenings." The connection of the rod to the cross head at 
one end and to the crank at the other is of such a kind as to allow 
perfect freedom of adjustment in the " 'plane of motion of the rod," 
while very little^ if any, is permitted in that at right angles to the 
plane of motion. 

In theory, therefore, the bending of the rod in the two planes 
should be treated as two different cases. The formula we have 
assumed applies directly to bu.ckling in the plane of motion of the 
rod. It can be applied also to determine the strength against 

buckling in the other plane by substituting -— for 1, on the ground 

that the strength of a column " fixed at both ends " is equal to 
that of one of half its length " free at both ends." One other 
change must be made. The radius of g3^ration employed is always 
that for the section of the rod about an axis in the plane of the 
section perpendicular to the plane in which lending takes place. 

Since all other factors are unaltered, if the value of W in the 
second application of the formula is not to change, the alteration 

in k must counterbalance that in 1. If k changes to -^ this con- 
dition is evidently fulfilled. Therefore, to give equal strength 
against both linds of huchling the radius of gyration of the section 
about an axis in the plane of motion must equal half the radius of 
gyration about an axis perpe^idicular to the plane of motion. 

For any circular, or flattened circular section, solid or hollow, 
likely to be used for a connecting rod, the two,rf^adii of gyration 
will be so nearly equal that we shall have a great excess of strength 
against bending sidewise. 

For a rectangular section the two radii are proportional to the 
sides to which they are parallel, in other w^ords, to the sides per- 
pendicular to the axes considered. If it is taken as a rule that 
the thickness of the section at right angles to the plane of motion 
shall equal or exceed half that in the plane of motion, bending 
sidewise will again be provided against. 

Thus no double application of the formula is ordinarily required 
and the only k considered is that for bending in the plane of mo- 
tion of the rod. 

63. Tapering the Rod. — The rod does not need to be of equal 
section throughout its length. Were it really a column pure and 
simple it could be made to taper in both directions from the middle. 



THE THEORY OF THE CONNECTI^-G ROD. 137 

The section at any point along the rod at a distance x from one end 
can be obtained as follows : The piece of rod of length x forms a col- 
umn " fixed at one end " (i. e.^ fixed or made fast by the molecular 
forces which bind it to the other part of the rod) and " free at 
the other." Such a column is equivalent to one of twice its length 
" free at both ends." The section at a distance x from one end 
is found, therefore, by putting 2x in place of 1 in our formula. 

If the sections at a great number of points are determined the 
resulting column will be one swelling gradually at the middle 
like an ornamental architectural column. Eods were at first so 
designed. More frequently the sections at two points near the 
ends were determined and the rod made to taper uniformly from 
the middle in both directions. Many slow-moving rods are still so 
made. 

With the increase of speed of engines it has been found desirable 
to increase the strength of the rods near the crank end. This is 
due to the additional bending stresses set up by the " transverse 
inertia " of the connecting rod itself. These stresses increase 
rapidly with the speed of the engine and are a maximum in the 
part of the rod near the crank end where the cross motion is 
greatest. 

The tapering of the rod at the crank end was first given up, 
leaving that portion of the rod of parallel section. Finally the 
practice has come in of carrying on the taper, or rate of increase, 
which is determined by the sections at the cross-head neck and 
the middle of the rod, to the crank end. This practice will be 
followed in the succeeding pages. 

64. The Rod Considered as a Tie Rod. — The formula for the 

C 
strength of a tie rod or short column is W = ^^ X A. This expression 

is the numerator in the formula for the strength of the rod as a 
column, equation (102), and we note also that the denominator is 
necessarily greater than unity. W will, therefore, always exceed W, 
and the rod designed for strength as a column will be amply strong 
as a tie rod. 



CHAPTEE XI. 
Peactical Problem III.* 

TO DESIGX A CONNECTING ROD WITH STRAP END FOR A SLOW- 
MOVING ENGINE. 

65. Specifications and Data. — Compound engine. Diameter of 
H. P. cylinder, 30 inches. Stroke, 36 inches. Boiler pressure, 80 
pounds. Connecting rod of wrought iron. Length between centers, 
90 inches. Dimensions of crank pin, 8 J" x 10". Assumed ratio of 
length to diameter of connecting rod, 15. Assume a solid circular 
section for the rod: C = 36,000, N = 9, E = 30,000,000. 

Gordon^s formula is 

W= ^PT- 



^^TT^E^ k2 

d^ 
Since k^ for a solid circular section equals — tt the formula can be 

16 

written 

Trd^ 



W= 



N X 



^"'" _2T?. X 12 



16C 
^E^ d^ 
or (at middle of rod) 

d=-^ \ "^ "'. (103) t 

* From " Notes on Machine Design." 

P 
fin equation (103) tlie factor^ , is often written r^ where r is the 

assumed ratio of length to diameter. The factor Ji^ is also written 

7r2E 



as 4a and \ -^ X -^ , as b and 103 becomes 

^^ yW(l + 4ar J2 (104) 

b 
When formula (104) is used the values of a and b must be first calcu- 
lated for the material to be used. This simplified formula is convenient 
only when a number of different rods of the same material are to be 
designed. 



PRACTICAL PROBLEM III. 



1:39 



The formula ( 103 ) is solved by assuming a value of ^ from a knowl- 
edge of similar designs ; having found the value of d, the true value 



I 5eciioN on XY 

I 




-f-f-ir- 

Q. O 



..t.,,_4._.4...i-_-...-_. 



H 




J 



.._.^A^ 



IX 




Fig. 84. 



of the ratio -V- can be found and the result must agree reasonably 
d 

well with the ratio assumed. If not found to agree, a new trial value 

of -r is taken and a second solution made, 
d 



140 EXGTlsTEEEIi^TG MECHA]STICS. 

Application of formida to find the middle diameter of the con- 
necting rod. 

E, the thrust on the piston rod, is, by the Bureau's method, 706.9 
X 80 = 56,552 pounds. ^ = sin-i i = ll° 31'. sec^ = 1.021.* 

W = E sec ^ = 56,552x1.021 = 57,700 pounds. 

Substituting in (103), 



J 57 700 fi+ _16X36,000_ 

V ^^^^^1^^^ 10X30,000,000 ^^^ , 

'^" /36£00 T ~^'^^- 

V 9 X 4 

This would make the ratio -^ = 17.7, or nearly 18, instead of 

15 as taken. A second application of the formula, with—^ =18, 
gives 



16X36,000 



y' 57,700 { 1 + .^r/o;r;:Arx^^ x is^ 



, . 10X30,000 ,000 ^^ 

736,000 TV '' 

V 9 ^ 4 
The diameter of the rod at the cross-head end may be taken as 
.9d = .9x5if=4ir. 

The diameter of the rod at the crank-pin end may be taken as 

l.ld = 63V'- 

Brasses (empirical dimensions, Unwin). — The dimensions of the 
crank pin, obtained from other calculations, are 8^ inches diameter 
and 10 inches long. The unit for calculating the brasses is found 
from t=.08d + ^", where d is the diameter of the pin. 

Then unit = t = .08x8r5 + f'=r80. 

B, the thickness at the bottom, = 2t=l'' 6 = li|'\ 

C, the thickness at the edges, = .7t='.'56=yV. 

D, the overhang, = 2it = 2.5 x '.'8 = 2". 

E, the width of the flange to hold the brass in the strap, = 

f _ 13" 

I — -j-g . 

F, the thickness of the same, = t = yf''. 

* It is difficult to pick accurately from the omnimeter the secants of 
small angles. This can be avoided by using Fig. 82 and working out 
the problem geometrically. In this example, replace 96" by 90" and 
24'' by 18", then 

90 



^90^—182 



= 1.02. 



PRACTICAL PROBLEM Hi. 



141 



At G, the brasses are planed away at the edges, as shown, in 
order to reduce the labor of fitting the brasses. 

The contour of the line joining the end of the overhang and the 
end of the flange is empirical. Xo. 5 is generally nsed in naval 
designs, as it is amply strong and is lighter than all except Xo. 1, 





(J] 



Fig. 85. 




4 




which is seldom used. In this case the curves at outside and in- 
side are about as drawn in Xo. 4. 

Stiih End and Strap. 

XoTE. — In calculating the dimensions of the stub end, strap, 
etc., tt is taken as 4000 and f., as 3000 pounds per square inch. 

H, the width of the strap. = the width of the stub end = length 
of crank pin — 2 x overhang = 10" — 4" = 6". 

I, the breadth of the stub end, = diameter of pin + 2 X thickness 
of the brasses at the edges = 8^^2 Xi^" = 9t". 

The thickness of the strap is found by calculating the dimensions 
to resist a tensile stress. Then, 

K, the thickness of the strap, x width xff=iW, or, Kx 6x4000 
=i X oT.rOO. Whence E:= l':20, or l^^", nearly. 

If the calculated value of K is ever less than the value taken for 
E, then either K must be increased or E diminished, so that K is 
greater than E by at least y^-". It is usually better to increase K. 

Key and Gib. — These are given a thickness of about one-fourth 
the width of the strap. 

L, the thickness of the key and gib,=-|" = l|-". 

Then, since the gib and key need be calculated only for shear, 

M (breadth of key and gib) xL (thickness) xfs^^W, or, 
M X If X 3000 = 4x57,700, and M = 6'.'41, or, 6^1". 

Erom Eig. 78 : 

X=: breadth of key = fMr=:f x 6-y-'' = 2r40, or, 2i|". 
0= breadth of gib = M-X 
10 



"3 2 



1 3 " _ A" 

^3T — ■* • 



142 



ENGINEERING MECHANICS. 



The taper of the key is -J inch to the foot. 

The key is made of such length that the small end projects from 
■J inch to 1 inch and the large end from 1 inch to 3 inches past the 
ends of the gib, depending on the size of the connecting rod. 

The thickness and length of the hooked ends of the gib are, 
as shown in Fig. 78, viz, 0.5b and 0.4b, where b is the combined 
breadth of key and gib. 

It is necessary for this size of connecting rod to allow ^ inch in 
the length of the slots in the stub end and strap for drawing down 
the key in adjusting brasses. So P, the length of the slots in this 
case, is 6^" + i" = 6^". The dimensions of the slot are, then, 

erxir. ^ 

strap. — Since one-fourth of the strap has been cut out for the 
key way, it is necessary to thicken it at this point in order to keep 
the necessary strength. Hence 

Q, the thickness of the ends of the strap, =|-xly6"" = l''^S, or 



^T2 ' 

This enlargement of the thickness begins a short distance above 
the key at E, and is carried to the ends. 

Crown of Strap. — To allow for extra bending stresses, the thick- 
ness of the crown, at S, is made 1.2 times the thickness of the 
strap, or 

S = 1.2 X IrV = 1-42, or l^f ". 





-1 

1 

1 

1 
1 
1 
1 






1 ^- i 






^ 


L.-'- ';..-'-'■ ^ 



3000 



Fig. 86. 
X2UX9J 



The distance T of the key way 
from the end of the strap is calcu- 
lated for shearing, the metal being 
in double shear. Hence 

3000 X T X 2 X lit " = 4 X 57,700, 
andT = 3r02,or33V'. 

The distance U of the key way 
from the end of the stub end is cal- 
culated in the same way : 
57,700, and IJ==1". 



In case the value of U in this calculation ever falls below \ inch 
it should, nevertheless, be made at least \ inch to provide for shock 
and for slight flaws in the material. 



PEACTICx^L PROBLEM III. 



143 



To find the curves of intersection of the rectangular stub end 
with the surface of revolution formed by sweeping the assumed 
curve of the enlargement of the rod around the central axis : 
The curve at a is assumed. Make a front elevation b of the stub end. 
Pass cylinders with centers at c longitudinally through the rod. 
These will cut the surface swept by a in circles concentric with the 
axis, the projections of which on d will be straight lines. They also 




Fig. 87. 



cut the sides of the rectangle. The intersections of these projec- 
tions give the points of the curves V and W. 

The cylinder e cuts the curved figure at a circle whose radius is 
cf. Project f to g and draw gh, the projection of the circle cut by 
the cylinder, e also cuts the rectangle at the point j. Project j to 
k and one point is determined. 1 and m are projected to n and o for 
the top and bottom of the curve, W, in the side projection. The 
remaining points of W are found by using the vertical side of the 
rectangle (front elevation) in the same manner. 



144 ENGIXEERI^^G MECHANICS. 

QUESTI0>^S AND PROBLEMS. 

State the three ways in which a connecting rod may fail. Which 
one is it necessary to consider, and why ? In the f ormnla 

W = 





C 

N 


XA 






a 




p 


-L + 


n' 


X- 


k^ 



explain the meaning of the several terms and show how the maxi- 
mum load W is obtained. 

Explain why a connecting rod of circular section is stronger as 
regards bending in the direction parallel to the shaft than in the 
direction perpendicular to it. What considerations govern the taper 
to be given a connecting rod ? 

Make a neat sketch of the stub end of a strap connecting rod, 
showing in detail all the features, including the provisions to com- 
pensate for the wear of the brasses. 

Problems. 

1. Find the diameter at the neck of a connecting rod from the 
following data: Diameter H. P. cylinder, 35 inches; stroke, 38 
inches; boiler pressure, 100 pounds; length between centers, 95 
inches. Assume a trial ratio of length to diameter of 15. Solid 
wrought-iron rod. 

2. A connecting rod is subject to a maximum load of 60,000 
pounds; the width of the strap is 7 inches. Find the dimensions 
of the key and gib and of the strap. See Fig. 84, P, 0, N, L, K, 
Q, M, S and U. Assume 1 = 10 inches; ft = 4:000; fs = 3000. 

3. From the following data find the diameter at the middle and 
at the neck of a connecting rod. Diameter H. P. cylinder = 12 
inches; stroke=16 inches; boiler pressure, 100 pounds per gauge; 
length between centers, 40 inches; C = 40,000; N' = 8„6; E = 30,- 
000,000; fi = 5000; fs = 4000. Assume l/d = 15. Solid circular 
section. 

4. A connecting rod is subject to a maximum load of 11,550 
pounds. The width of the strap is 3 inches. Find the dimen- 
sions of the key and gib and of the strap. See Fig. 84, P, 0, N, L, 
K, Q, M, S and U. Assume I = 41 inches ; f ^ = 5000 ; f ^ = 4000. 



PRACTICAL PROBLEM III. 



145 



s s 



O '* 



§ g 



o^' 



6C 05 



a >< 

51 






?. b ^ i 

,-1 ,-1 cs «o 

>i X X X 



© fl o 



>* CO 









-^ ec! 



•ON 'qojd: 



-H N 



« -<*< 



146 ENGINEERING MECHANICS. 

Sequence of Calculations for the Connecting -Bod Prohlem. 

I. 

1. rind the maximum load on the rod. 

2. Find the middle diameter of the rod using the assumed ratio 
of length to diameter. 

3. Check the ratio of length to diameter from 2 and if necessary 
repeat the calculation. 

4. Find the diameter at the neck of the rod. 

5. Find the unit t for the calculations for the brasses. 

6. Calculate the dimensions of the brasses, B, C, D, E and F of 
Fig. 84. 

11. 

7. Calculate H and I, Fig. 84. 

8. Calculate the thickness of the strap, K, Fig. 84. 

9. Calculate L, M, N" and 0, Fig. 84, and the length and depth 
of the hooked end of the gib, Fig. 78. 

10. Find P, Fig. 84. 

HI. 

11. Calculate in order Q, S, T and U, Fig. 84. 



CHAPTEE XII. 



Practical Peoblem IV.* 

67. To Design a Connecting Bod for a Naval Vessel. Specifica- 
tions. — A method of designing a connecting rod of a type much used 
at present for naval vessels is illustrated in detail in the following 
pages by giving the calculations for a rod for the battleship Ala- 
bama, as an example. 

Before beginning the design, certain main dimensions and speci- 
fications are decided upon from general considerations of the type 
of engine and from the design of the adjacent parts. 

For the Alabama they are as follows : 



Name of ship, 

Number of screws, 

Type of engines, 

I. H. P., 

Diameter of cylinders, 

Stroke, 

Revolutions, 

Boiler pressure, 

Material of connecting rod. 

Ultimate strength of material, 

Factor of safety for rod, 

Length between centers. 

Dimensions of crank pin, 

Dimensions of cross-head pin, 

Depth of fork from center of 

cross-head pin. 
Cross-head pin faced to, 

68. The Load the Rod Must Bear. — A general idea of the type of 
rod to be designed and the names of the parts can be gained from 
Fig. 88. The sections to be first determined are those at aa and bb. 
For them Gordon's formula is used, and as a preliminary the value 
of W must be determined from the specifications. Thus 

R, the thrust on piston rod = area x pressure = 8814 x 180 
= 158,612 pounds, or 158,600. 



Alabama. 

2. 

Vertical triple expansion. 

10,000. 

331", 51'^^ 78". 

48". 

120 per minute. 

180 pounds. 

Forged nickel-steel, oil-tempered. 

80,000 pounds per square inch. 

8.5. 

96". 

143" diameter X 17". 

9f" diameter X 14". 5" bore hole. 

: 19". 



* From " Notes on Machine Design. 



148 



E2vrGINEERING MECHANICS. 



^ = sin-i-||=:sin-ii=:14° 29'. Sec^=1.033. Whence W = E sec ^ 
= 158,600x1.033 = 163,800 pounds. 

68a. Application of Gordons Formula. — C and N are given in the 
"specifications" as 80,000 pounds and 8.5 respectively. E = 30,- 
000,000 pounds. tt'^IO. Thus we have 



W = 163,800 



80.000 
8.5 



XA 



1 + 



80,000 



10x30,000,000 



X 



96^ 



K2 



(105) 




cu-ffrng Vsr\e^ ^honiet line 







Plan stub 

_ eHd 




f3m 



— main iot<g ing 

^"^^ Q .b 




^middle t> |en^f/i— j 



'cr^ead 31^ EJeva+ion 




View 



distance Legend 



piece 



Fig. 88. 



A and K^, our unknowns, are not independent. If made to de- 
pend on common variables (as d^ and do below) the equation is of 
high degree and not easily managed. It is best to solve simply by 
trial and error. 

Since we are confined to hollow sections (b or d in Fig. 89) be- 
cause of the oil-tempering, our first trial will be a circular section 
of 8 inches diameter with a 5-inch bore hole. The diameter of the 
bore is usually from .5 to .6 of the outside diameter. We will 
calculate proper values for A and k from these dimensions and sub- 
stitute in the right-hand member of the equation. The value of W 
found, will then be compared with the value required, 163,800 
pounds. 



PRACTICAL PEOBLEM IV. 



149 



Calling the diameter, area, and radius of gyration of the 8-inch 
circle, d^, A^ and k^, and those of the 5-inch circle, ds, A^ and ks, 
then A=Ai — Ag. 

Eeferring to the table of areas of circles, we find A = 50.27 — 19.64 
= 30.63 square inches. 

The value of k- in terms of d^ and da can be derived from the 
fact that the moment of inertia of the hollow section is equal to 
the moment of inertia of the solid 8-inch circle — the moment of in- 
ertia of the 5-inch circle. Expressed mathematically, I^I^ — Ig, or 

k^A^ki^A.-k^^A^. Thenk^ = ^^'f^~^^'^^ 




solid hoKow 

circular, circular. 



m 



^ 



'A 



in- 



e. 



f 

solid hollow K^cf double 

laliened flatfened aJoijIar ^^t" 
circular, circular.. >3^ angvlar. 

Usual Sections of Connecting Rods. 

Pig. 89. 



these substitutions, 

k2^ d,^-d,^ 



di2, and A, 



Making 



16(d,^-d,^) 



d.^ + d. 
16 



A 2 A 2 

16 "^ 16 



A table of values of -— has been calculated and is given in the 

back of the book. 

From it we find k2 = 4.000-f 1.563 = 5.563. Then, 

80,000 ..on^o 

9,410x30.63 



W: 



>.0 



80,000 



■^"^ 10x30,000,000 
288,200 



X 



96- 



1.442 



5.503 
199,900 pounds. 



1 + 



2.4574 
5.563 



As this valne exceeds 163,800 pounds by a considerable amount we 
make a second trial reducing each diameter by -J inch. Then 



150 ENGIN'EEKING MECHANICS. 

di=:7i", d2 = 4i", A = 44:.18-15.90 = 28.28, k2 = 3.516 + 1.266 

.1 ^Qo ;i w 9410x28.28 266,100 -,^. q^^ . 

= 4.782, and W= — ; — ^ . j — = ' =175,800 pounds. 

^"^ 4.782 
This value is almost exactly two-thirds of the way from 199,900 
to 163,800 pounds. Eedncing again by ^ inch we may expect a very 
close result. Thus, di = 7i", d- = 4i", A = 41.28-14.19 = 27.09, k^ 

Q oQPc , 1 -lOQ I M ( A w 9,410x27.09 254,900 

= 3.285 + 1.129 = 4.414, and W= 2.4574 = 1.557 ^ 

■'-+ 4.414 " 
163,700 pounds, a solution as close as need be. 

The diameter of rod and bore hole are, then, 7Jinches and 4J 
inches. 

69. Section at the Neclc. — The neck is assumed to be at about one- 
third the length of the rod from the cross-head end, say 30 inches. 
The true position of the neck will be determined later. The section 
of the rod at 30 inches from the free end is theoretically the middle 
section of a rod 60 inches long. The bore hole, 4J inches in diameter, 
is of course retained and at first trial the outside diameter may be 
supposed to decrease about ^ inch. Then, di = 6|", d^ = 4J", 
A = 21.59, K- = 3.977, from which 

9410x21.59 203,200 .ror^r^f^ a 

^= ., 80,000 00- = "IMT =^^^'^^^ P^^^^^' 

10x30,000,000 ^ 3.977 

a very good result, so that no further trial is needed. 

The value 203,200 pounds, is the strength of the rod as a tie rod, 
since it is the strength of the smallest section. 

70. The Fori-. — The fork is faced at the sides to a thickness 
greater than the diameter at the neck by from 5 to 25;^. In this case 
we may choose 8 inches (A in Fig. 90). The width between the 
jaws is -| inch or J inch greater than the specified length of the 
cross-head pin. This allows a clearance, or room for a slight motion 
of the rod. The distance B (Fig. 90) is, therefore, 14 J inches. The 
distance C must be such that the fork of the rod may clear by a large 
margin (1 inch to 6 inches) the cap and bolts of the cross-head 
brasses in every position that the rod may take relative to the cross 
head. It must be obtained from the drawing of the cross head, as 
shown in Fig. 91, or must be given in the data prepared by a pre- 



PRACTICAL PROBLEM IV. 



151 



vious designer. In this case C is 19 inches. The radius D is, of 
course, J B. 

The thickness E must be such that the section of the fork is 
from 70 to 100;^ of the rod at the neck. Large rods require the 
larger amount as the tempering of the thick metal (over 2^ inches), 




^^r—- ^ ^ T 




Sec+ion 

■ 


of F 


'ork 


r 




\..,_ 


^^Y 






-^.y 


r 




""n^ 




m/ 
1 




^ 


V 

\ 
1 




jt. 


-A- 


1 

1 
/ 
/ 

/ 




^- 


^ 


L_^-.. 




iU 




. > 






1 















Sect 



ion on iLiL 



Fig. 90. 



is imperfect. It is supposed that much more than half the load may 
come on one side of the fork from defective alignment. Taking 
95fc in this case we have .95 x 21.63 = 20.55 square inches. Since the 

20.55 



8 



width of the section is 8 inches the average height must be 

= 2''57 = 2yV'- If ^^e take E = 3" it will allow amply for the round- 
ing down of the corners of the section when the outer surface of the 
fork is turned in the lathe, the rod being centered on its axis. (In 
Fig. 90 the dotted arcs, m and n, complete the circle traced out 
by the tool in cutting these surfaces.) 



152 ENGINEERING MECHANICS. 

The radius F is equal to D + E = 7|" + 3" = 10J". The longi- 
tudinal contour of the fork is continued to the right of the parallel 
portion (a) by running into the circular arc (b) of the same ra- 
dius F. The center of this arc should be from ^ inch to 2^ inches 
to the right of the center of D. This strengthens the fork at the 
curved part, and it is still further strengthened where it joins the 
middle length of the rod by rounding in with a large radius, H, 
about equal to F. In this case G = 2" and H = 10". 

The distance J is measured on the accurate drawing of the rod. 
It is the true distance of the neck from the center of the cross- 
head pin. If it is less than the assumed 30 inches the error is on the 
safe side and no recalculation is needed. In this case J = 28f". 

71. Cross-head Pin. — The bearing length of the pin, 14 inches, 
has been specified to suit the design of the cross-head brasses. The 
diameter K^ has been found by allowing a pressure of 1200 pounds 
per square inch of projected area. Thus 14 x9|x 1200 = 163,800 
pounds. The bore hole Kg lightens the pin, and enables it to be 
oil-tempered. The pin is of the same steel as the rod and is shrunk 
or forced into the eyes of the fork. A hole is then drilled and tap- 
ped half in the pin and half in the eye, as shown at P. A " screw 
plug " or " stud " is screwed tight in the hole and the head cut off 
flush with the surface. This plug prevents the pin from working 
loose and turning in the fork. For a small rod it may be as small 
as i'' diameter X f " long. For a large rod two plugs are used as here 
shown. These are 1^" diameter x 2" long. 

The diameter L is slightly larger than K^ to allow for a small 
flUet at the end of the bearing length of the pin. In large and 
medium rods this increase is -J inch and ^ inch. Small rods do not 
require it. In addition M is greater than L by from -^ inch to yg- 
inch. The small end L of the pin will then slip easily through the 
large bore M of the fork in the process of inserting the pin. In this 
ease L = 9|" and M = 9i|-". 

The eye of the fork is reinforced by making the depth N" greater 
than E, and the diameter grekter than A. The area left in the 
plane qq after boring the hole M is equal to from 90;^ to 100^ 
of the area of the neck. This rule is empirical. The exact strains 
cannot be calculated owing to the unknown bursting strains due 
to the pin. Choosing 95fo we get (0 — M)N' = 20.55 square inches. 
Assuming = 14i", N = 4^". 

The arc j is usually given a radius equal to A. 



PRACTICAL PROBLEM IV. 



153 



The arc k has a radius equal to or less than N" — E. In this case it 
is 1^ inches. 

Most of the bearing length of the pin is faced at the two sides 
to a width Q somewhat less than K^. The sides are useless as 
bearing surfaces since the force acting is always in line with the 
axis of the rod. By cutting them away a pocket is formed for 
oil to collect in, and to act as a kind of reservoir. A small length 
at each end, say J inch, is left circular to help retain this oil. 

Fig. 91 is a sketch of the cross head with the fork end of the 
rod shown at the extreme angle of swing. The cross-head pin and 
brasses are sectioned and the near side of the fork removed. The 



/4-'29' 




Fig. 91. 



width Q is so adjusted as to allow the edge, x, of the bearing sur- 
face of the pin to slightly over-travel the bearing surface of the 
brass it works against. All bearing surfaces whose action is re- 
ciprocating should be designed with this over-travel at the point of 
extreme displacement. It prevents a shoulder or step being formed 
in the brass by the wear not extending to the edge. Without 
going into the design of the cross-head brasses we shall assume that 
we know the distance Q and that it is in this case 8J inches. 

72. The Size of Bolts, Crank End. — The general shape of the 
crank-pin brasses is seen in Fig. 93. The cap is held by two bolts, 
made of the same steel as the body of the rod. We imagine that f 
of the whole load on the rod may come on one bolt through unequal 
tightening of the nuts. The bolt is never in compression, but its 
load fluctuates with each revolution of the engine from the designed 



154 



ENGINEEEING MECHANICS. 



tension to zero tension. A lower factor of safety can be used than 
for the rod itself in which the fluctuation is from tension to 
compression. 

For torpedo boats and destroyers (thrust in connecting rod, to 
100,000 pounds) the factor averages 5.8 in the designs of our naval 
vessels. For cruisers and battleships (thrust 130,000 to 245,000 
pounds) it averages 7.8. For gunboats (thrust to 40,000 pounds), 
8.5. 




Nut ^^R 




Solid Bolt 




Hollow Bolt 



KL-J 



Fig. 92. 



In the present case we will use a factor 7.5. Then for the area 
of the bolt at the weakest section we have 

Areax §^5^5 z=|x 163,800. .•. area= 10.22 square inches. 

The two styles of bolts in general use are shown in Fig. 92. In one 
design the solid bolt is turned down in two places so that the area 
remaining is about equal to the area of the threaded part of the bot- 
tom of the threads. The bolt thus reduced and made of fairly uni- 
form strength along its length is stronger to resist sudden shocks 
than if left of full size. It is more yielding, more " resilient.^^ By 
referring to the " Table of Standard Bolts and Nuts for the U. S. 
Navy," on p. I, and the " Table of Areas of Circles " on p. II. 
the dimensions, A = 4", C = 3f", will be found suitable for our 
case. The area at B is thus 10.61 square inches, and C it is 10.33 
square inches. 



PRACTICAL PROBLEM IV. 155 

The second design accomplishes the purpose of making the bolt 
of uniform strength along its length by means of the bore hole E. 
The 4-inch bolt has a section of 12.57 square inches or 2.34 square 
inches more than the 10.22 square inches required. A bore hole, E, 
of If inches diameter, will remove 2.07 square inches, leaving the 
area at A equal to 10.50 square inches. 

For convenience in oil-tempering the bolt, the large bore hole 
E is usually continued by a small one, F, through to the end. 
If F = f" the metal remaining at B will be 10. 61 -.31 = 10.80 
square inches. 

Be3^ond the nut the bolt is turned to the diameter G=:C = 3f" 
and is extended far enough to allow a round split pin to be passed 
through the hole T. Split pins vary from -J inch to f inch in 
diameter, depending on the size of the bolt. In this case the diam- 
eter is -J- inch. The pin should bear against the face of the nut when 
in position. 

The projecting end of the bolt is J inch long for the J-inch split 
pin. 

The length of the bolt and -the positions of the turned down 
places cannot be determined until the brasses and cap have been 
designed. The threaded length of the bolt should exceed the depth 
of the nut by from J inch to 1 inch. Thus 

D = P-F0 + r = 5H". 

73. Eye-holt Holes and Set Screius. — At U a hole is drilled and 
tapped for a 1-inch eye bolt. The depth of the hole is 1^ inches. 
Medium-sized bolts are tapped for f-inch eye bolts, IJ inches deep. 
Set screws are fitted in the cap or in the brasses to hold the bolts 
after the nuts have been slackened and removed, until the eye bolts 
have been screwed into place and slings fitted. 'No groove is re- 
quired for these set screws as they always bear at one place which 
can be flattened with a file. Small bolts require no set screws or 
eye-bolt holes. 

Set screws are standard square-headed steel screws (see table on 
p. I). The head of the f-inch set screw here used has a short 
diameter of IJ inches and a long diameter of If inches. If neces- 
sary, it can be made much smaller than the standard. The point is 
turned down to ^ inch diameter. It projects J inch and is fiat 
across the end. 



156 ENGINEERIN^G MECHANICS. 

74. Bolt Heads and Nuts. — The bolt heads are round and of less 
dimensions than the standard heads^ as no twisting force is ever 
brought upon them. They are held from turning by dowel pins or 
'^ snugs." For the solid bolt^ which has an excess of strength in the 
section just under the head^ the snug is rectangular and is set into 
the metal of the holt. In this case it is f inch square, set in ^ inch 
and projecting -J inch. For the hollow bolt it is a pin set in the 
metal of the bolt head by drilling completely through the head. For 
this bolt it may be f inch in diameter and f inch longer than the 
depth, L, of the bolt head. 

The under surface of the bolt head is in compression. This 
area should equal the section of the bolt. The whole area of the 
bolt head is, therefore, 10.23 + 12.57 = 22.80 square inches. This 
will be given by a diameter, H, of 5^ inches. 

The thickness of the bolt head, L, is fA = |x4" = 3". 

The nut is of wrought iron, case-hardened on the outside. The 
short and long diameters, M and N", are taken from the Table of 
Standards. They are 6J inches and T^^g- inches for a 4-inch bolt. 

The bearing surface of the nut is the bottom of the nut. The 
bottom of the hexagonal portion stands clear of the metal by yV inch, 
when the nut is screwed home. The diameter J is a little larger 
than H, in this case 5f inches. The groove for the set screw should 
be yV inch or ^ inch deep. Without the groove the burr raised by 
the pressure of the set screw would injure the fit of the nut. Thus 
we take Kr=5J". The height of the hexagonal part of the nut is 
f of the diameter of the bolt. = 3". The height P is from J 
to f of 0. The groove E is in this case f inch wide to allow for a 
f-inch set screw. We will take Q = -i" and S = yV'5 "tt^^s making 

75. Position of Center Lines of Bolts. — The inside edges of the 
bolts should clear the crank pin by from J inch to f inch, according 
to the size of the crank pin. In this case f inch is sufficient. The 
distance from center of crank pin to center of bolt is, therefore, 

crank pin 5,, bolt diameter ^ 14|^ 1 5'/ , o"_io" 
2 "^ 8 "^ 2 2 ^' ~^ ~ • 

76. The Stub End of the Main Forging. — The width of the 
" stub end " (parallel to the crank pin) is from 60fo to 80^ of the 
length of the brasses. The crank pin is 17 inches long (see Speci- 
fications in Art. 67). T, the corresponding length of the brasses 



PRACTICAL PROBLEM IV 



157 



(Fig. 93), is 16f inches, to allow the same side clearance, i inch, as 
is allowed for the cross-head pin. Taking 80^ we find D = .80 
Xl6:'75=:13r4 = 13i". 

The distance of the butt from the center of the crank pin must 
now be determined. The thickness of the brass at Q is about J 



hK^ 



^5-1 



;>], I,, J 




Hn^- 



Distance Piece Removed 




h- --D-H 

End Elevation 




^zz 



Distance Piece 



White metal 



Fig. 93. 



the radius of the crank pin. Thus Q = ^ x 14$'' = 2f ' nearly. The 
distance R is equal to 2^' + 7f " = 9|". 

The width, E, of the stub end, is equal to the distance apart of 
the bolts + the long diameter of the nut + a small amount for clear- 
ance. In this case let ^ = 20" + 7-^^" -\- jY = 27i'\ This allows 
^2" at each end for the corners of the nuts to keep within the edges 
of the stub end. 
11 



158 ENGINEERING MECHANICS. 

Tlie thickness G must allow siifficient metal for strength after 
recessing the seat for the nut. If G is made equal to the diameter 
of the bolt (4 inches), the proportion will be a good one. The dis- 
tance H is yV' i^ch less than P in Fig. 91. Thus H = If". 

The radius C is determined when the laying down of the de- 
sign on the drawing board has reached this point. The arc must 
be tangent to the tapered section of the rod and also to the sur- 
face of the stub end at a point, a, at or within the center line of 
the bolt. It is usual to . so adjust this radius as to make F, the 
diameter of the rod at the point of tangency, some integral number 
of inches and eighths, for convenience in the shop. If necessary, 
the diameter at the middle section, bb, in Fig. 88, is slightly in- 
creased to allow this. In this case, C may be made 5| inches and 
F, 8J inches. The same radius, C, appears twice on the side eleva- 
tion, the centers being on the same vertical line as the centers in the 
plan. 

In addition to the recess for the circular part of the nut there 
is a recess to allow room for the corners of the hexagon as the nut 
turns. This recess is nearly a semicircle. Its diameter is slightly 
larger than the long diameter of the nut. In this case it is 7J 
inches, or JrrSyV. 

The end surfaces of the stub end are cylindrical. In the end 
elevation they appear as arcs of a circle whose center is 0. 

77. The Cap. — The cap is of oil-tempered nickel steel and is 
regarded as a beam supported at the ends and loaded with a uni- 
formly distributed load. In other words, the bolts are not counted 
on to keep the ends of the cap from assuming a slight angle with 
their unstrained position. Undoubtedly they do resist such a ten- 
dency, but the strengthening effect gained may be offset by the fact 
that the load is in reality distributed over the middle three-quarters 
of the length of the beam only. (The length of the beam is the 
distance between the centers of the bolts.) 

The greatest bending moment is at the middle, therefore, and is 

., Wl 163,800X20 .r^o Knn • T A 

equal to -^— = ^-^ =409,500 inch-pounds. 

o o 

The moment of resistance of the rectangular section at the mid- 
die is: f^x ^. (b = the width, D, of the cap, and h = the 



PRACTICAL PROBLEM IV. 159 

thickness L.) Using the same factor of safety as for the rod, 8.5, 

it, the safe working stress, = — ^' »- = 9410 pounds. 

8. o 

The equation for h, or L, becomes, therefore : 

409,500 = 9410x W^ , or h^= ^JM. =19.34, 

whence h = 4'/4. We shall increase this to 4'.'5. 

The thickness of the cap at the ends, M, is about f L. In this 
case 3 J inches. For small caps, M = L. 

Large caps have the bolt heads sunk in a distance N, a little 
less than H. For our cap, let N" = l^ inches. The diameter of the 
recess is -J inch larger than the diameter of the bolt head. Thus 
= 5f inches. A recess receives the snug on the bolt. 

P is equal to E. When the bolt heads are not recessed. P=: 
distance apart of the bolts + diameter of bolt head + J''. 

The set screw which holds the bolt when the nut is slackened 
may set to press against the bolt head or the bolt itself, as is most 
convenient. 

78. The Crank-pill Brasses. — The dimensions Q and E have been 
determined. 

S, the distance taken up by the " distance piece '' and tin liners, 
varies greatly with the individual designer. A good value for a 
14j-inch pin is 4 inches. Of this amount J inch is taken up by the 
tin liners. 

The two large brasses are alike, save for the boss Y on one, 
which fits into the bore hole of the rod. The thickness U of the 
brasses at the overhang is about 10</o of the length of the crank 
pin. In this case, we take it as 1-J inches as the unsupported " lip " 
of the brasses is small. The outer surface of the lip is conical, the 
taper being 15° on each side. 

Most of the other visible surfaces of the brasses are obtained by 
keeping the thickness as constant as possible. In this case a thick- 
ness, W, of 1 inch, is taken, at the places shown in Fig. 93. 

Frequently tap bolts are used to fasten the " cap brass " to the 
cap and the " rod brass " to the stub end. Four are used for each, 
and are so placed as to utilize the waste material in the corners 
of the cap and rod. They are here shown at jj on the end view 
and ]']' on the plan, and are 1 inch in diameter. The holes in the 
cap and forging are drilled and tapped 1^ inches deep. 



160 ENGINEERING MECHANICS. 

Dowel pins projecting from the cap brass serve to keep the dis- 
tance pieces and liners in place. Two pins are used for each dis- 
tance piece. The pins may be -|" diameter x 2" long. The holes 
in the brass are i" X 1" deep and are a '^ driving fit." The holes in 
the distance pieces are |-" + 1-J'' deep and are a " working fit." 

The distance pieces are of horse-shoe shape, for convenience in 
removing them. The cap bolts need only be slackened back enough 
to free the distance pieces from the dowel pins. The diameter, V, 
of the concave semicircle is equal to the bolt diameter + yV. 

79. White Metal Lining. — The brasses and the ends of the dis- 
tance pieces are lined with white, or anti-friction, metal. The 
finished thickness ranges from y\ inch to -^q inch with the size of 
the rod. Taking it as f inch, the inside diameter, x, of the nnlined 
brasses becomes 14|"4-f" = 15J". In addition, many grooves, ^" 
deep X J" wide, with undercut edges, are spaced at equal angles 
around the circle. Two at least must be put in each distance piece. 
These grooves hold the white metal firmly in place. 

The middle of the distance pieces and adjacent parts of the 
brasses are not lined, thus leaving a shallow cavity (f inch deep) in 
which oil collects and from which it is redistributed over the crank 
pin. 

The white metal, when cast over the concave surfaces of the 
brasses, is made considerably thicker than the finished work. It 
is hammered down to a dense condition with a riveting hammer. 
The parts having been assembled and bolted together, the metal 
is bored out and scraped to an exact fit with the crank pin. 

80. Oil Holes and Ducts.— The provision for oiling the crank 
pin is usually given in detail on a separate drawing. Two copper 
tubes of f inch outside diameter, one running down each side of the 
rod, lead the oil from the cross-head end, where it is received drop 
by drop in a narrow funnel-shaped opening, to holes bored in the 
main forging along the lines 11. The holes extend nearly to the 
brass with the diameter of | inch and are continued through the 
brass and white metal with a smaller diameter. 

From the openings on the surface of the white metal (see points 
mm on the end elevation. Fig. 93), oil ducts are cut by a grooving 
chisel leading in some pattern intended to spread the oil over the 
surface of the crank pin. 

Sometimes the oil is introduced down the center line of the rod 
by a tube inside the bore hole. In such a case the hole through 



PRACTICAL PROBLEM IV. 



161 



the brass and white metal is in the center of the boss Y and the 
oil ducts radiate from the point o in the end elevation. 

Alternative Designs. 

81. Solid Circular Middle Section. — This section is now little 
used in naval vessels, as it is not a favorable form for oil-tempering. 
It may be used for gunboats. 

Since k, the radius of gyration of a disc about a diameter, is 
one-fourth the diameter, the formula can be written 



W= 



X 



d= 



16C 



(106) 



^^ TT^E ^ d^ 



and solved by trial and error, but the application of the general 
formula, making do = 0, is j ast as simple. 

82. Flattened Circular Sections. — These sections (c and d of 
Fig. 89) are modifications of the sections a and b. The regular 




Fig. 94. 



solution for the section a or b is first made including the size at the 
neck. Usually the section at the neck is not flattened at all and the 
middle section is flattened at the sides to a width equal to the outside 
diameter at the neck. (Of course the whole middle length is then 
faced to this width.) The flattening diminishes the area of the 
section, and to compensate, the outside diameter is increased. 
Graphical tables of the radii of gyration of flattened circular sec- 
tions, solid and hollow, have been constructed and by their use the 
exact increase can be determined. 



162 



ENGINEERING MECHANICS. 



A fair solution, of which the error is on the safe side, is obtained 
by making the added area equal to that removed. Fig. 94 shows 
by this method the determination of the increase in the diameter of 
the rod of the Alabama at the middle, if it is desired to face the 
sides to the diameter at the neck. The area removed is approxi- 
mately f xthe chord a (which is by measurement 2f") xthe altitude 
b, or f x2f"xi'' = .437 square inch. By stepping along the arc 
c with the bow spacer set to J inch we find the arc equal to 8f inches. 

487 

The increase required is, therefore, -^-^ ='.'05. The new outside 



8.75 



diameter is, therefore, 7'.' 3 5. 



--"-^ 




Fig. 95. 



83. Rectangular Section. — Values for the width and thickness of 
the section are assumed and the formula used to test them in the 
regular trial and error method. The values of t and w must be such 



w^ 



thatw^2t. Since A = wxt, and K^ = -y^, the formula becomes 



W 



XtXw 






(107) 



'E 



The use of it is illustrated in the following example : 

For a torpedo boat, given W = 52,000 pounds, 1= 40", C = 80,000 
pounds, ISr = 8.5, assume first that w = 3|", t=:2". Then W = 



PRACTICAL PROBLEM IV. 



163 



46,500 pounds. Assume next, w = 3J", t = 2. Then W = 51,700 
pounds, a good solution. 

The thickness t is generally kept constant and the width w made 
to taper uniformly from the neck to the crank end. Assuming 
the neck to be at 13 inches from the cross-head pin, we find that, if 
w = 3f", and t=:2", W will equal 53,400 pounds. This results in a 
taper of f inch in 7 inches, an unusual amount, changed with advan- 
tage to ^ inch in one foot. This change will increase the width at 
the middle by yV ii^ch, i. e., to 3 |f inches. 

84. Open Fork, Cross-head End. — Large rods are sometimes de- 
signed with the jaws of the forked end fitted with small " marine 



ion on 




Fig. 96. 



ends,^^ reduced copies of the crank end of the rod. In such cases the 
cross-head pin is shrunk in the cross head at the middle of its length, 
the two projecting ends forming end-journals for the brasses on the 
jaws. It is supposed that f of the load may come on one jaw from 
irregularity in fitting, and the ends are designed for that load. The 
end of each jaw is forged out to the shape shown in Fig. 95, to give 
room for the bolts. White metal is often omitted from these brasses. 
The nuts of the cap bolts bear on the caps, not the fork. 

" Solid " Cranh End. — The crank end and cap are sometimes 
designed in the manner known as a ^' solid end.^^ The brasses are 
reduced to thin semicircular pieces held in the steel by " nipping 
bolts." The end of the main forging and the inner surface of the 
cap are concave semi-cylindrical surfaces instead of flat surfaces. 



164 ENGINEERING MECHANICS. 

To reduce the distance apart of the cap bolts, they are made to pass 
partly through the brasses. Fig. 96 shows this design. The section 
on ab is that through the nipping bolts. 

85. Cap with Bounded. Ends. — The cap and the similar portion 
of the main forging are often finished off with semicircular ends, 
instead of ends formed by turning them on the center line of the 
rod. Thus in Fig. 93 we should have instead of the arc d"d', whose 
center is o, a semicircle of diameter D, passing through the point 
d'. The two ends of the fork in Fig. 95 have been rounded in this 
manner. 

86. Brass Cap. — Small rods very frequently have a. brass cap in 
lieu of a steel one. The cap in such cases takes the form of the 
steel cap in Fig. 96 combined in one piece with the cap brass. The 
rest of the design may follow either Fig. 93 or Fig. 96. In calcu- 
lating the thickness of the brass cap^ a low-working fiber stress is 
taken, its value being about half that for the steel. " Brass " caps 
are generally made of some form of bronze. 

Original Calculations and Sketches for a Connecting Rod. 

87. Note Booh, Sketches, etc. — Sets of specifications like those 
in Art. 67 will be given out in the drawing room. From them each 
one will make his own design of a connecting rod for a naval vessel, 
following in a general way the methods given in the pages just pre- 
ceding, but depending on his own judgment for a selection from 
among the alternatives mentioned for the different details. 

These specifications are the first entry on the interleaved pages and 
the calculations, as they are made, are neatly entered in conjunc- 
tion with free-hand sketches of the details of the rod to which they 
refer. The sketches are of detached parts of the whole, like Figs. 
90, 92 and 93, or of even smaller portions. Dimension letters are 
not used, but the values as determined by the calculations are re- 
corded on the sketches which then become the basis from which the 
drawing of the rod is ultimately made. 

88. Order of Recordirig Calculations. — The following table gives 
the order in which the required dimensions are classified and re- 
corded. It is, as far as possible, that in which they are calculated 
or determined: 

In sketch books place dimensions on the left. 



PRACTICAL PROBLEM IV. 165 



I. Specifications (see Art. 67). 

II. Middle Length. 

1. Thrust on piston rod. 
crank 



2. Sec ■{ sin 



connecting rod 

3. Thrust on connecting rod. 

4. Factor of safety of connecting rod. 

5. Type of cross section, middle. 

6. Dimensions of middle section. 

7. A, K^, and W for middle section. 

8. Assumed position of neck. 

9. Dimensions of section at the neck. 

10. A, K^, and W for section at the neck. 

11. Strength of the rod as a tie rod. 

12. Diameter of rod at cross-head end (to be filled later). 

13. Corrected size at middle section (to be filled later). 

14. True distance of neck from center of cross-head pin (tc 

be filled later) . 

III. Fork End. 

15. Width of fork in plane of motion of rod. 

16. Width between jaws of fork. 

17. Area of fork required. 

18. Thickness of fork at center. 

19. Distance between centers for inside and outside circular 

contours of fork. 

20. Radius for rounding in at the neck. 

21. Diameter of cross-head pin within top eye of fork. 

22. Diameter of cross-head pin within bottom eye of fork. 

23. No. and size of screw plugs for fastening cross-head pin 

in eyes of fork. 

24. Outside diameter of eye of fork. 

25. Depth or thickness of eye of fork. 

26. Eadius for rounding in side lines of fork and outside 

diameter of eye of fork. 

IV. Cap Bolts, etc. 

27. Factor of safety of cap bolts. 

28. Area of cap bolts. 



166 ENGINEERING MECHANICS. 



fDiameter and area of turned- down part of bolt. 



29. ^ or 



(^ Diameter and area of bore hole in bolt. 

30. Diameter and area of small continnation of bore hole. 

31. Diameter of bolt beyond nut. 

32. Length of bolt beyond nut. 

33. Diameter of hole for split pin. 

34. Threaded length of bolt (filled later on)., 

35. Total length of bolt under head (filled later on). 

'Position of turned-down places (filled later on). 

36. J or 

Depth of large bore hole in bolt (filled later on) . 

37. Diameter and depth of eye-bolt holes. 

38. Diameter of bolt head. 

39. Depth or thickness of bolt head. 

40. Dimensions of snug. 

41. Short and long diameters of nut. 

42. Depth of hexagonal part of nut. 

43. Depth of C3dindrical part of nut. 

44. Width and depth of groove on nut. 

45. Width of collars on each side of groove. 

46. Size of set screws. 

47. Size of heads of set screws, if not standard, 

48. Diameter of point of set screws. 

49. Length of point of set screws. 

50. Length of set screws under head (to be filled later). 

51. Distance apart of bolts. 

Y. Stul) End and Cap. 

52. Width of stub end parallel to crank pin. 

53. Width of stub end perpendicular to crank pin. 

54. Thickness of stub end parallel to center line of rod. 

55. Depth of recess for cylindrical part of nut. 

56. Diameter of recess for hexagonal part of nut. 

57. Eadius of arc of enlargement by which middle length 

joins stub end. 

58. Thickness of cap at center. 

59. Thickness of cap at ends. 

60. Depth of recess for bolt heads. 

61. Diameter of recess for bolt heads. 

62. Length of cap over all, perpendicular to crank pin. 



PRACTICAL PROBLEM IV. 1G7 

VI. Brasses, etc. 

63. Length of crank pin covered by brasses. 

64. Length of overhang, or lip of brasses. 

65. Thickness of brasses at overhang. 

QQ. Thickness of brasses between crank pin and stub end. 

67. Distance from center of crank pin to stub end. 

68. Thickness of distance pieces. 

69. Thickness of tin liners. 

70. Thickness of brasses over bolt. 

71. Thickness of flanges. 

72. Size and position of tap bolts fastening brasses to forg- 

ings. 

73. Size and position of dowel pins for aligning distance 

pieces. 

74. Width of horse-shoe opening in distance pieces. 

75. Thickness of white metal lining. 

76. Depth and width of nndercnt grooves for holding white 

metal. 

77. Number and angular position of grooves. 

78. Dimensions of cavity where white metal is not placed. 

79. Diameter and direction of oil holes through main forg- 

ing. 

80. Diameter of oil holes through brass and white metal. 

81. Pattern and size of oil ducts or grooves. 

The different groups into which these dimensions have been col- 
lected should have their own sketch or sketches to illustrate them 
and to show the exact details. 

Some of the dimensions are here entered in advance of their 
natural order of determination. Such are left blank for the mo- 
ment. A few are determined only on the accurate drawing of the 
connecting rod. With the notes otherwise complete proceed to lay 
down the design on the drawing board, but finally return to the 
notes and fill in those dimensions here marked " to be filled later.^^ 

The Working Drawing of the Connecting Eod. 

89. Arrangement of Views and Sections. — The figures in the 
preceding pages illustrate special points and are not models to be 
followed in the drawing. 

The connecting rod should be shown with all yarts assembled. 



168 ENGINEEKING MECHANICS. 

For a vertical engine the plan, in an architectural sense, would be 
an end view of the rod from the fork end. Practically the rod 
is considered as lying on one side, and that view in which the 
cross-head and crank pins show as circles, is taken as the "plan.'* 
The (other) side elevation is placed under the "plan." Contrary 
to the general rule it is advisable to put the end view or elevation 
(looking on the cranh end) on a line with the " plan/^ not with the 
" side elevation." This arrangement is shown on a small scale in 
Pig. 88. It makes a compact drawing, as the end view is narrowest 
in the direction parallel to the crank pin and it leaves a space con- 
venient for the legend. 

When necessary to show the interior structure, parts of the plan 
and elevations are made " in section," as in Figs. 93 and 96. (In 
Fig. 93 the bolts are absent, as is often the case in sTcetches, In the 
drawing they should be in place, as in Fig. 96. 

Often small additional sections are drawn, without showing any 
part beyond the plane of the section itself, like the sections of the 
cross-head pin and the fork in Fig. 90. These are intended to 
show clearly to the eye the strength of the metal at those places. 

90. The Scale of the Drawing. — ^The scale depends on the size 
of drawing space to be devoted to the rod. The cutting, border and 
working lines are first laid down to the sizes specified in tl:ie drawing 
room. The actual distances, c and d, on the connecting rod (see 
Fig. 88) are computed and added together and the result compared 
with the horizontal clear space, e, between the two side-working 
lines. Such a scale is chosen as will make the apparent size of 
c + d a little less than e. The distance left over must provide first 
for a good separation between plan and end elevation, and then, if 
still sufficient, for additional space at the ends like f and g in Fig. 88. 

91. Center Lines, Circles, etc. — The vertical center line through 
the cross-head pin is first laid out by measuring from the left-hand 
working line. From it is measured the center line of the crank pin. 
The vertical center line of the end view is measured from the right- 
hand working line. The horizontal center line of the plan is meas- 
ured down from the top working line. The center line of the side 
elevation can be put in by eye alone unless the vertical height avail- 
able is scanty. The center lines of the bolts are then drawn. The 
circles representing the cross-head and crank pins on the plan, the 
dotted circles for the diameters of the rod on the end view, the arcs 
for the ends of the cap and the full and dotted circles for the diam- 



PRACTICAL PROBLEM IV. 169 

eters of bolt heads and bolts are then drawn. The circles for the 
contour of the fork inside and out follow. As a general rule, circles 
are drawn first. Very often b}^ projecting tangent lines from these 
circles from one view to another double measurements of the same 
distances are saved. 

The points aa and bb are now established. Their horizontal 
distances from the cross-head pin are measured off, and their ver- 
tical positions projected from the end view. The rod as finally de- 
signed may not follow these points exactly, but the contour must 
include them. 

x\ll the fundamental dimensions so far drawn should be verified 
and altered until absolutely correct. 

The details of the ends of the rod are now taken up in much 
the same order as that of performing the calculation. 

92. Some of the lines on the drawing are obtained, not by meas- 
ured distances, but by a geometrical construction. Thus, in the plan 
in Fig. 93, the line marked d'" is derived by projection from the 
point d" in the end view. 

Many of these geometrical lines are of one class and represent 
the appearance of the edge or intersection of a surface of revolu- 
tion and a plane parallel to its axis. Such curves are a'", b'", c' 
in Fig. 93, c and d in Fig. 96, and fgh in Fig. 90. 

The method by which points are found on such a curve is illus- 
trated in Fig. 93, by finding the points b"'. The " surface of revo- 
lution " is in this case the " flared '^ or '" bell-shaped " surface by 
which the middle length of the rod is enlarged to join the stub 
end. The plane is the flat side of the stub end. The contour 
of the flared part is formed, both on the plan and on the side 
elevation, by arcs of the saine radius C, whose centers all lie on 
the same vertical line. 

One point on the edge, c', can be taken at once by projecting up 
from c on the side elevation. If any plane is passed perpendicular 
to the axis of the flared surface at a point between c' and h, it 
will cut out a circle, bb is such a plane and it appears as a line 
on the plan, while on the end elevation the circle it cuts out is 
seen in its true shape. A part of the circle is the dotted arc 
b'b"b", whose center is o, of which the point b' is taken by pro- 
jection from b on the plan. The two points, b", lie both on this 
circle and on the flat edge of the stub end. Project from b" to the 
plane bb on the plan. The points b'" so marked are the same 



170 ENGINEERING MECHANICS. 

two points and are, therefore, on the required edge between the 
two given surfaces. 

The plane ha is the limiting plane to the right, the plane to 
which the bell-shaped surface becomes tangent. In the same way, 
therefore, a, the point of tangency of the arc of radius C, gives 
rise to the arc a'a"a", and the points a" give the points a'" on the 
line ha. Outside of the points a'" the curve is straight. 

If the dimensions F and D are more nearly equal than in Fig. 
93, the curve becomes a more marked one, and it may be well to 
find more points than those here found, a'", b'" and c'. When 
enough points have been found, by passing new perpendicular 
planes, the curve can be drawn in with irregular curves, or arcs of 
circles found to approximate to it. 

When the curve is as fiat as in this case, since nothing depends 
on the accurate drawing of it, determine only the points a'" and 
c'. Pass through them three arcs of circles tangent to each other, 
finding the centers by trial. The center of the middle one will be 
on the center line of the figure to the right of c', the centers of 
the others to the left on horizontal lines through the points a. 

The curve c in Fig. 96 is like that just described. 

The curve d (Fig. 96) arises when the entire surface of the cap 
visible on the end view is turned on the lathe with the rod cen- 
tered on end centers. Both c and d can be represented by three 
arcs of circles. 

The curve fgh in Fig. 90 is of the same kind. The sectional 
view in Fig. 90 is not needed on the completed drawing, but some 
portions of it must be drawn in pencil for use in finding the curve 
fgh. A convenient place for it is at x in Fig. 88. 

In Fig. 90 the outer surface of the fork between the planes rr and 
ss is cylindrical, and in consequence the edge f is straight and is 
given by projecting from the corner f in the sectional view. Be- 
tween the planes ss and tt the surface is truly spherical. The edge 
g is, therefore, a circular arc whose center is 0. Between tt and uu 
the surface is bell-shaped. The point h" is projected down from 
h' on the plan. Intermediate points can be found by passing 
planes between tt and h", but the edge h is usually represented by 
three circular arcs tangent to each other and to the arc g, the 
middle arc passing through h". 

The line f (Fig. 90) is continued to the left by the line 1, which 
meets the upper surface of the fork at a point projected from the 



PRACTICAL PROBLEM IV. 171 

plan, as shown. The exact drawing of the curve is immaterial 
It is -usually drawn a circular arc and need not continue quite to 
the upper surface of the fork (see Fig. 88). The line represents in 
reality the edge produced by the hand filing necessary to finish 
neatly the cutting out of the corners at j. 

In Fig. 93 the recess cut in the stub end for the corners of the nut 
to turn in^ is almost a semicircle. The line of the edge appears on 
the side elevation approximately as the half ellipse ge'g, in which 
the major axis gg is equal to 2xJ (on the plan) and the point e', 
the extremity of the minor axis, is projected from e on the plan. 
Any approximate method of drawing the ellipse by circular arcs 
will be sufficiently accurate. 

93. Marking Finished Surfaces. — A working drawing should 
show definitely which surfaces are machined and which are left in 
a comparatively rough condition ("rough f orged ^^ or " cast ^' sur- 
faces). The general rule is to mark every planed, turned or slotted 
surface with an "f (meaning "finished") on that part of the 
drawing where it is seen on edge and is represented by a line. The 
f is put right on the line it refers to. The letter " b " or the word 
" bore " is used to distinguish holes bored or drilled from those 
made by a core in the process of casting. It is put after the dimen- 
sion figures giving the diameter of the bore hole thus |-<— f b— >|. 

In most connecting rods, the brasses, which are castings, alone 
have unfinished surfaces, and it is customary, therefore, to put on 
the drawing in a prominent place a note to this effect, " All sur- 
faces finished except on brasses^ which are finished where marked.'^ 
The brasses, in Fig. 93, have been marked in conformity with this 
practice to show where to put the " f s." 



CHAPTER XIII. 

Shafts, Journals. 

torsion, strength of shafts. twist of shafts. size op shaft 
per given horse-power. strength of journals and pins, 
combined turning and bending moments. couplings and 

BOLTS. 

94. The General Theory of Torsion. — Fig. 97 represents two 
shafts, B and C, fitted with couplings and connected together by 
the pin A, the two couplings being very close together, the shaft B 
driving the shaft C, the power being the pnll W acting at the 
distance 1 from the axis. It is evident that the pin is in shear, and 



^ 



' B 



\ 



Fig. 97. — Torsion 




if we call F = the shearing resistance of the pin, we have, by taking 
moments about the axis, 

Wl = Fy. (108) 

Let a=:area of the pin. 

fs = shearing stress of the pin. Then 

Wl^f^ay. (109) 

Now let ns have two driving pins between the couplings ; then 



Wl: 



(110) 



Fy + F^yi r=f say + f sialyl. 

The dotted circles in the figure represent the pin holes in the driv- 
ing coupling when the rotating force W is acting. Before W is 
applied the holes are exactly in line, but when the force is applied 
the yielding of the pins allows a slight movement of the couplings 
relative to each other. 



SHAFTS^ JOURNALS. 



17;] 



This yielding, or the strain^ varies directly as the distance of the 
pin from the axis, and since the material is supposed to be elastic 
the stress varies directly as the strain, and we have 

Y = ^ ort,,= ^^; 

U y y 

substituting this value of fg^ in equation (110) we have 



(111) 



Wl = f„aY+ ^^ = ^(ay^ + a,y,^) 



(112) 



Suppose now instead of only two pins we connect the couplings 
with a very great number of very small pins, so great a number as 
to form a solid section between the couplings. Let the areas of each 




Fig. 98. — Torsion. 



of these small pins be a, a^, a^, etc., and their distances from the axis 
be y, Y-i, Jo, etc., respectively. Then we have 
f. 



Wl: 



( ay 2 + a^y^^ + a,y/ + etc. ) . 



(113) 



The quantity in the parenthesis is the product of each very small 
area by the square of its distance from the axis; in other words, it 
is the polar moment of inertia of the section. This is usually de- 
noted by the symbol I^. Then 



Wl 



(114) 



I. 



but -^^ mav be called the polar modulus of the section which is 

y 

denoted by Z^. Hence 



W] = isZ,. 



(115) 



12 



174 ENGIXEERIXG MECHAXICS. 

The moment Wl is called the twisiing or turning moment, and is 
denoted by T, so we have 

T = isZp, (116) 

which is similar to equation (18) for bending moment. 

In this case, however, ts is the shear stress of the material fur- 
thest from the axis and Zp is the pola?- modulus of the section in dis- 
tinction from the modulus for bending. 

95. The Strength of Circular Shafts to Resist Torsion.^FTom. 
equation (114) above it is seen that the strength of a shaft depends 

I 7rd^ 

upon the value of — ^— == Zp, but Ip — , where d is the diameter 

y ^^ 

of a solid circular section (see Art. 44) , and y = — ^^ ; hence we have 

Trd^ 

This is just twice the value of Z for bending. It is easy to recol- 
lect which value to apply by considering the fact that in a circular 
shaft the metal is disposed in the very best manner to resist torsion, 
but is in a bad form for bending; hence the torsion modulus will 
be greater than that for bending. 

For a hollow shaft d external, and d^ internal diameter 
(d^_d,4) 
^~ 32 ' 

Z - 32 _ ,r(d^-d,^) _ / d^-d,^ 

^v- ^ - ^^ -0.196' 
2 

96. The Twisting of Shafts. — Eeferring back to Chapter III, Art. 
23, it was shown that the amount of slide x when a piece of material 
was under the action of a shearing force was determined by the 
relation 

1 G ^ ^ 

fs = the shearing stress of the material. 
G=the coefficient of rigidity. 



SHAFTS, JOURNALS. 



175 



To apply this to the case of a shaft under a twisting action, the 
slide X is measured around the circumference, as indicated in Fig. 
99, and it is usually expressed in terms of the twist angle, thus : 
The circumference of the shaft =:7rd. 

The arc which subtends 1° at the axis=: oarf* 
The arc which subtends 0° at the axis=: ^--r- 



= x. 




Fig. 99. — Twist. 



Substituting this value of x in equation (117) gives 

rrdO 

3C01 
But from equation (116) we have 



U a 360f,l 



T — f sZp — 



16 



whence 



f.= 



Trd^ 



Substituting this value of f^ in equation (118) gives 
360x l 6xTl _ 584T1 

^~ TT^Gd^ Gd^ 



(118) 



(119) 
(120) 



1^6 E^^GIJsTEEEING MECHA^^ICS. 

for a solid circular shaft. Similarly, for a hollow circular shaft, 

0- 584T1 

The above twisting formulge are used when it is desired to keep 
the spring of a shaft within a given amount, the usual limiting 
value of the spring or twist being 1° in a length equal to 20 diam- 
eters. Thus from equation (118) 
._ 360fsl 

when 

0=1°, l = 20d; 
then 

' 360x20d 2292 * ^ ^ 

From equation (122) Ave get the following: 

For steel, G= 13,000,000; fs = 5670 lbs. per sq. in. 

For wrought iron, G = 11,000,000; fs = 4800 lbs. per sq. in. 
For cast iron, G= 6,000,000; fs = 2620 lbs. per sq. in. 

These values of f,? are those that mu.st be used to keep the twist not 
greater than 1° in 20 diameters, and show that for long shafts the 
diameter will be considerably greater than is necessary for mere 
strength. 

When the shaft is short, so that the consideration of spring or 
twist does not have to be considered, the following values of the 
working stress may be used : 

Mckel steel, f « = 12,000 lbs. per sq. in. 

Steel, fs = 10,000lbs. persq. in. 

Wrought iron, fs= 8,000 lbs. per sq. in. 

Cast iron, f « = 3,000 lbs. per sq. in. 

97. The Size of Shaft to Transmit a Given Horse-Power. — The 

problem that is most frequently presented is to find the diameter of 

shaft of a given material necessary to transmit a given horse-power. 

The following method is probably the clearest and most direct for 

the solution of this problem. 

Suppose the shaft is being revolved against a resistance by a 
force of F * pounds acting at a distance 1 inches from its axis ; then 

* In a steam engine the force P and the turning moment are not 
uniform throughout the revolution, but the method gives the mean 
turning moment. The maximum turning moment, in naval engines is 
usually from 1.25 to 1.5 of the mean, according to the type of engine. 



SHAFTS. JO CECALS. 177 

F (pounds) xl (inches) = twisting moment on the shaft; in inch- 
pounds =T. 

In each revohition the force F acts through a distance of 27rl 

9 1 

(inches) or -^ (feet) ; hence 

Fxlx27r 



work done per revolution in foot-pounds, 
number of revolutions per minute 
work done per minute in foot-pounds, 



12 

If 1^ = number of revolutions per minute 

Fxlx27rX]Sr 

12 
but 33^000 foot-pounds per minute = 1 H. P. ; hence 

-n- -D , ., , -, Flx27r]N" 

H. p. transmitted =j^^^33^^, 

or 

F1 = T= H. P. X 33,000x12 ^ ,^^3^ 



From equation (116) we have 



T = f.Z„=-^-; (124) 

hence 

fsTrd' ^ H. P. X 33,000x12 
16 27r:N' 

or . 

.3_H. P. X 33,000x12x16 ,,^.. 

In all reference books this formula is found given in the form 

where C is a constant depending on the value of fs. This is really 
the same as formula (125), the constant 



p 733,000x12x16 

^=V 2.-^f, 

98. Journals and Pins. — The journal is that part of a machine 
which rotates in the hearing^ the latter being the stationary part. 
In some cases the motion of the journal in its bearing is occasional 
only, and then the size of the journal must be calculated principally 
for strength. In other cases the motion of rotation is continuous, 



17 



ENGINEERING MECHANICS. 



r\ 



and in these cases, in addition to strength, considerations of heating 
are equally important. Again, in some cases the load acts on the 
journal in such a manner as to produce shearing or bending only; 
and in other cases the journal is subject to torsional stresses at the 
same time as the bending stresses, when the size must be calculated 

by the method of combined 
P stresses. Fig. 100 represents a 

journal under a load P uniformly 
distributed over its length 1. Let 
d = diameter of journal and f = al- 
— lowable working stress of the ma- 
terial. Then the bending moment 
at the fixed end of the journal is 
JP1 = M. 

From equation (18) 

M = fZ = f^2 ^ 
hence 



Vj 



1 



■^i 



a 



^^ 



4P1 



/ 



/ 



or 



Pl = 



32 



Trd^ 
16 



f, 



f. 



(126) 



/ 



Fig. 100. — Journal. 



Suitable values of f are : 

Steel, f = 9000 to 12,000. 

Wrought iron, f = 6000 to 9,000. 

Cast iron, f = 3000 to 4,500. 

Unwin gives the following graphic method of finding the bending 

action : 

" Suppose, first of all, that the load on a journal (Fig. 100) is P 
acting at the center of its length. Then the bending moment 
diagram is the triangle abc, in which bc = ^Pl, is the bending mo- 
ment at the root of the journal. If the load is uniformly dis- 
tributed the bending moment curve is the parabola dc, which may be 
approximated to by drawing ce perpendicular to ac and describing 
a circular arc dc with center e. The bending moment at the root 
of the journal is the same as before, and on this the diameter neces- 
sary for strength depends." 

In most cases the value of 1, the length of the journal, is not 
known, and, therefore, there are two unknown quantities, 1 and d, 



SHAFTS; .lOFRXALS. 179 

ill equation (126). However, it is oenerally possible to fix on a 

ratio of-y , the following being custoinarv values: 

For journals woi'king intermittently, or on which the pressure 

acts for a part of a revolution onl\ . , = 1 : for journals working^ at 

■ (1 ■ 

speeds below about 150 revolutions ])er minute, =1.5 to 1.25 

for composition '^brasses.*' or— =1.15 for cast-iron bearings. 
With a fixed ratio of 1 to d^ equation (126) reduces as follows: 

16 ' 

dividing through by d gives 

1 - 12 

r = ' f, 

(I IG ' 
or 



V^^R)- <-) 



This formula is used for finding the dimensions of an over-hung 
crank pin such as is found on the ice machine on board ship for 
strejigth. 

]Many cross-head pins are fixed solidly into the forks of the con- 
necting rod, and are then in the condition of uniformly loaded 
beams, fixed at the ends. If, then, as before, P = the total load, 
1 = length of the pin and d = its diameter, we have 

PI 



also 



whence 



''- 12 ' 



M = fZ = f.'^; 



PI, _r Trd^ 

12 ~ 32 



Pl_ 

Trf ' 



(128) 



or when the ratio -p is fixed, 
d 



cp 1 

d==3f^x;, (129) 



180 ENGINEERING MECHANICS. 

99. In calculating the dimensions of journals of all kinds it is 
not sufficient to consider only the torsional and bending stresses to 
which they are subject due to the load, but in addition the intensity 
of the surface pressure under which they act must be taken into 
consideration, and also the rubbing speed of the bearing surfaces. 
This is usually done in practice by fixing a maximum pressure per 
square inch of projected surface which must not be exceeded. The 
dimensions are first calculated for strength and are then checked to 
make sure that they come within the allowed pressure per square 
inch of projected area. 

The projected area is merely the area obtained by the product 
1 X d of the bearing. 

The following limiting pressures per square inch of projected 
area, taken mostly from Unwin, are very generally accepted : 

TABLE 17. Pressure 

persq. in. of 
bearing sur- 
face iu lbs. 
Bearings carrying intermittent loads at low speeds, such 

as pins of shearing machine 3000 

Cross-head neck journals; motion oscillation; not complete 

rotation (1000 commonly used) 800 to 1400 

Crank pins (slow engines) 800 to 900 

Crank pins (fast engines) 500 to 800 

Crank pins (marine engines) 400 to 500 

Crank pins (small land engines) 150 to 200 

Crank shaft bearings (slow) 300 to 450 

Crank shaft bearings (fast) 200 to 300 

Eccentric sheaves 80 to 100 

Thrust collars (slow) 70 to 80 

Thrust collars ( fast) 50 to 70 

Cross-head slides (Babbitt on cast iron) 200 to 300 

Steel or iron shaft lignum vitas (under water) 350 

100. Combined Turning and Bending Moments. Equivalent 
Turning and Bending Moment. — In many shafts, particularly the; 
crank shaft of a marine engine, there is not only the torsion due to 
the horse-power transmitted, but there are, in addition, severe bend- 
ing stresses due to the thrust of the connecting rod on the crank 
pins. 

In such cases there is a shearing stress due to torsion and a 
tensile stress due to the bending action. In Art. 50 it is shown that 



SHAFTS; JOURNALS. 181 

the simultaneous action of shearing and tension produces a maxi- 
mum stress^ Pma« = i-f^ + Vifi" + fs^, see equation (61). To apply 
this formula to the case of a shaft of circular cross section, 

Let T = twisting moment at any given section. 
M = bending moment at the same section. 
We have now to find the equivalent twisting moment Tg or the 
equivalent bending moment Mg, that is to say, the single simple 
twisting or bending moment which would produce an effect on the 
shaft equivalent as regards strength, to that which is actually pro- 
duced by the twisting and bending acting together. 

But 

T — is'Zip, where Zp is the torsion modulus; 

f - ^^ = A^T 

M = ffZ, where Z is the bending modulus; 



. M 32M 



Z Trd- 



Also 



P,„«. = if. + Vf/ + W. 
Then the equivalent twisting moment is 



; — J^ max X Zip, 

=z.Kf i + VfT+W] = ^ [4f < + VfT+W], 

16 



fi 32M , /7T6TV, , /32MV1 



TT • ^-(M+V1- + 1P) 



or 



Te=M+VT2 + M^ (l;50 



Similarly 



Me=Pma^'XZ, 

=z(if, + Vf7+W), 

which reduces to 

Me =pi + iVT^ + M^. (131) 

The formula Me=fM + fVT^nF is sometimes given in text 
and reference books. This formula is arrived at by considering the 
greatest strain produced in distinction to the greatest stress, and i^ 
based on a value of Poisson's ratio of 0.25. See footnote, Art. 50. 



182 



ENGIXEERIXG MECHANICS. 



The common Am^eriean and British practice is to nse the form given 
in equation (131). In designing shafts the equivalent twisting 
moment, Te, is used rather than the equivalent bending moment, 
Me, but the use of either will give the same result. 

101. Shaft Coupling. — The several lengths of shafting are con- 
nected together by means of couplings and coupling bolts. When 
an engine has two or four cranks the complete crank shaft is usually 
in two sections, and the number of bolts in the coupling is then an 
even number. When three cranks are used the complete crank shaft 
is usually in three interchangeable sections, and the number of bolts 
in the couplings is then a multiple of three, six being the number 
commonly used. 




Fig. 101.— Shaft Coupling. 



Fig. 101 shows the usual proportions of a shaft coupling the unit 
d being the diameter of the solid equivalent shaft, such that 



V " d, 



To find the diameter 8 of the coupling bolts proceed as follows : 
Let H. P. be the horse-power transmitted. 

n be the number of bolts. 

K be the radius of the pitch circle. 

N be the number of revolutions per minute. 

A be the area of section of each bolt. 

fo be the shearino- streno-th of the material of the bolt. 



The turning moment on the shaft is, then, T = 



11. r.x 33,000x12 

" 27rN"' 



see Art. 97, equation (123). This turning moment produces a 



SHAFTS, JOURNALS. 183 

force F acting at the arm Iv, giving a moment = F x K ; this latter 
moment is equal to the turning moment. 

.'.T^FxK. (132) 

But this force F is a shearing force acting on the bolts, whose com- 
bined area is nxA. 

.■.nxAxfs = F. (133) 



Whence 



or 



T = nxAxf«xK, (134) 

T ■ 



A = 



nxf.xK' 



but 



A=-^^ 



4 
.'.8 



71111 

V ^nf.K ' 



/4xH. P. x 33,000x12 r.oK\ 

= \ 2.^Nnf,K ^^^^^ 

A simplified formula for finding the diameter of the bolts in terms 
of the diameter of the shaft will be given later. 

The outside diameter of the coupling is large enough so that the 
corners of the nuts of the bolts do not project beyond the flange. 
Usually 1.9 to 2.0d will be found a suitable value. 

After finding the diameter, as shown above with an assumed value 
of K of from .75d to .8d, it may be found that the corner of the nut 
will not turn clear of the surface of the shaft. This is not likely to 
occur, however, as it is the usual custom to turn down the threaded 
end of the bolt to less diameter than its general body. This can 
safely be done, as the bolt is in shear only, and the only function 
of the nut is to draw the sections of the shaft closely together. In 
the case of the line and propeller shafts, however, in backing the 
thrust of the propeller produces tension in the bolts, and conse- 
quently there is a force tending to strip the threads of the nuts, 
which must be taken into consideration. 

The diameter 8 is the bolt diameter at the joint between the two 
coupling disks. The bolts are given a taper of J inch on the diam- 
eter per foot length of bolt. 



184 exgixeeeixg mechanics. 

Questions and Problems. 

Explain the general theory of torsion. Dednce an expression for 
turning moment in terms of Zp and is- 

Deduce an expression for the twist angle of a shaft in terms of 
T, 1, G and d. 

Show how to obtain the limiting values of f .§ in order to keep the 
twist of the shaft within given limits^ say 1° in a length of 20 
diameters. 

Show how to deduce an expression for turning moment given 
I. H. P. and ^, the revolutions per minute; thence deduce an ex- 
pression for the diameter of shaft necessary to transmit a given 
H. P. in terms of H. P., revolutions per minute, and tg. 

What is the difference between a journal and a bearing. Deduce 
an expression for the diameter of an end, or over-hung journal in 
terms of load, length of journal and f. What are suitable values of 
f^ for steel, wrought iron, cast iron? Show how the bending 
moment may be shown graphically. 

Deduce an expression for the diameter of an end, or over-hung 
journal, in terms of load, length of journal, and f, when the ratio 
of 1 to d is fixed. 

Deduce an expression for the diameter of a " neck " journal, such 
as the cross-head pin when fixed in the forked end of a connecting 
rod. 

A shaft is subject to torsion and bending at the same time. What 
stresses are produced in the material? Given the expression, 
Pma:r = ift + Vif^"-t-fs", clcduce an expression for the equivalent 
turning moment Te in terms of T (turning moment) and M (bend- 
ing moment). 

Make a sketch of a shaft coupling and show how to find the diam- 
eter of the coupling bolts, to transmit a given horse-power. 

Problems. 

1. Find the diameter and length of the over-hung crank pin of 
an engine of the following principal dimensions : Diameter of 
cylinder = 45 inches; stroke = 42 inches; initial pressure =50 pounds 
per gauge; length of connecting rod = 12 feet; E. P. M. about 50; 
wrought-iron pin working in composition brasses. 



SHAFTS, JOUE^^ALS. 185 

jSTote. — ^lake t\ro calculations, first for strength, then check for 
maximnm allowable pressure from table (slow engine). Select 
ratio of 1 to d and value of f from text. 

2. Diameter of steam piston = 20 inches; steam pressure = 160 
pounds per gauge ; find the diameter of the cross-head pin, which is 
fixed in the forked end of the connecting rod. Select proper pres- 
sure per square inch of projected area of pin, ratio of 1 to d, etc., 
from text. 

3. Diameter of steam piston = 20 inclies; steam pressure = 160 
pounds per gauge; cross-head journal consists of two over-hung 
pins (usual naval type). Select proper data from text, and assume 
that f of the total load mav come on either pin. Find the dimen- 
sions of the pins and mark them on a sketch of the cross head. 

M 

4. Calculate a table, the first column to show values of -^ advanc- 
ing by tenths from 0.1 to 1.0; the second column to show the 
equivalent simple turning moment in terms of the turning moment 
T : the third column to show the equivalent simple bending moment 

M 

in terms of the bending moment M. (Suggestion: Let -^^ =x or 

:\r = Tx, and let ^ = - or T = Mv.) 
T y 

5. A solid steel shaft transmits 10,000 H. P. at 125 E. P. i\r. 
Find the mean turning moment and the diameter of the shaft. 

6. A solid shaft 12 feet long transmitting 300 H. P. at 200 
E. P. M., is loosely supported at each end. Weight of shaft is 520 
pounds (consider this weight as a distributed load producing bend- 
ing) . Find the greatest equivalent turning moment in inch-pounds. 

7. The I.H. P. of an engine = 5000; E. P. M. = 120; stroke = 48 
inches; distance between main bearings = 55 inches. Find (1) 
mean turning moment in inch-pounds; (2) maximum turning 
moment = 1.4 mean; (3) maximum turning force on after crank 

pin ( suggestion: force = ^j; (4) diameter of crank pin for 

bending strength, f^ = 10,000; (5) dimensions of crank pin, ratio 
1 to d = 1.14 and allowable pressure per square inch of projected 
area = 650 pounds; (6) from equivalent turning moment and 
fg=z8000, find diameter of after shaft journal; ratio of axial 



186 ENGINEERING MECHANICS. 

8. The pitch circle of the bolts of a flanged coupling has a diam- 
eter 11 times that of the shaft. I. H. P.r:r7000; E. P.M. = 130; 
fs=:8000; number of bolts = 8. Find the diameter of each bolt. 

9. The diameter of a shaft transmitting 5000 I. H. P. at 130 
E. P. M. is 16 inches, with an axial hole diameter = 8 inches. There 
are 8 bolts in the coupling, ts for bolts = 8000 pounds per square 
inch. Make a neat, fully dimensioned sketch of the coupling. 

10. A shaft, 4 inches diameter and 30 feet long, is found to 
twist 6.2° when transmitting power at 130 E. P. M. G = 12,000,000. 
What horse-power is being transmitted ? 



CHAPTEK XIV. 

Fi^iCTiON AND Lubrication of Bearings. 

GENERAL EFFECT. COMPARISON BETWEEN DRY AND LUBRICATED SUR- 
FACES. NOMINAL AREA OF BEARING. WORK OF FRICTION. VALUE 
OF /x. PRESSURE IN BEARING. FORCED LUBRICATION. ANTI- 
FRICTION METAL. THRUST BEARING. 

102. The General Effect of Friction and Lubrication. — The essen- 
tial condition for a properly working bearing is that it should 
run cool and not seize. It is impossible to have a frictionless jour- 
nal in practice^ and friction always produces heat, and, consequently, 
a rise of temperature. This rise of temperature, in well designed 
and carefully lubricated bearings, reaches a certain point at which 
the rate of generation of the heat by friction is equal to the rate at 
which the heat is conducted away, when the temperature remains 
constant. This temperature is the so-called "working heat," and 
so long as it is moderate the journal will work properly. The 
amount of friction is reduced to a minimum by supplying the bear- 
ing with a lubricant, which forms a film between the journal and 
its bearing. This not only greatly reduces the heat generated, but 
it also greatly reduces the amount of wear. 

If, from any cause, a rise of temperature above the normal occurs, 
the first effect is generally a reduction of friction by making the 
lubricant more fluid, but this reduction of the viscosity of the oil 
allows it to be squeezed out of the bearing by the pressure between 
the surfaces, and the friction then takes place between metal and 
metal, and seizing occurs. The heat now becomes so great as almost 
to weld the surfaces together, and pieces are cut out of both the 
journal and the brasses, causing deep scores in each. The load at 
which seizing occurs depends upon the smoothness of the surfaces, 
and the nature of tht3 materials, but mostly on the viscosity of the 
oil. If the viscosity can be kept up by artificial means, such as 
water circulation, very high pressures may be safely carried. A case 
is recorded by Dr. Goodman of a journal running continuously at a 
surface velocity of 230 feet per minute, with a load of two tons per 
square inch, the temperature 'oeiug kept at 110° F. 



188 



EXGIXEERIXG MECHAXiCS. 



The brasses of the main bearings of naval engines are made 
hollow, and are fitted for the circulation of water through them, 
which greatl}- assists in conducting away the heat of the friction, 
and thus prevents the oil from becoming too fluid. 

103. Comparison of Laws of Friciion 'between Dry Surfaces and 
Well-Luhricated Surfaces. (From G-oodman, " Mechanics Applied 
to Engineering/^) — The laws which express the behavior of well- 
lubricated surfaces are practically the reverse of those for dry sur- 
faces, as is shown by the following parallel columns : 



Dey Suefaces. 
1. The frictional resistance is 
nearly proportional to the normal 
pressure between the surfaces. 



2. The frictional resistance is 
nearly independent of the speed 
for low pressures. For high press- 
ures it tends to decrease as the 
speed increases. 



3. The frictional resistance is 
not greatly affected by the tem- 
perature. 



4. The frictional resistance de- 
pends largely upon the nature of 
the material of which the rubbing 
surfaces are composed. 



Lubricated Sukfaces. 

1. The frictional resistance is 
almoso independent of the press- 
ure with bath lubrication, and ap- 
proaches the behavior of dry sur- 
faces as the lubrication becomes 
more meager. 

2. T h e frictional resistance 
varies directly as the speed for 
low pressures. But for high press- 
ures the friction is very great at 
low velocities, becoming a mini- 
mum at about 100 ft. per minute 
and afterwards increases approxi- 
mately as the square root of the 
speed. 

3. The frictional resistance de- 
pends more upon the temperature 
than on any other condition — 
partly due to the variation in the 
viscosity of the oil, and partly to 
the fact that the diameter of the 
bearing increases with the rise of 
temperature more rapidly than 
the diameter of the shaft, and 
thereby relieves the bearing of 
side pressure. 

4. The frictional resistance with 
a flooded bearing depends but 
slightly upon the nature of the 
material of which the surfaces 
are composed, but as the lubrica- 
tion becomes meager, the friction 
follows much the same laws as in 
the case of dry surfaces. 



FRICTION AND LUBRICATIOX OF BRAKINGS. 



189 



Dry Surfaces. 
5. The friction of rest is slightly 
greater than the friction of 
motion. 



6. When the pressure between 
the surfaces becomes excessive, 
seizing occurs. 



7. The frictional resistance is 
greatest at first and rapidly de- 
creases with the time after the 
two surfaces are brought together, 
probably due to the polishing of 
the surfaces. 

8. The frictional resistance is 
always greater immediately after 
reversal of direction of sliding. 



Lubricated Surfaces. 

5. The friction of rest is enor- 
mously greater than the friction 
of motion, especially if thin lubri- 
cants are used, probably due to 
their being squeezed out when 
standing. 

6. When the pressure between 
the surfaces becomes excessive, 
which is at a much higher press- 
ure than with dry surfaces, the 
lubricant is squeezed out and 
seizing occurs. The pressure at 
which this occurs depends upon 
the viscosity of the lubricant. 

7. The frictional resistance is 
least at first, and rapidly increases 
with the time after the two sur- 
faces are brought together, prob- 
ably due to the partial squeezing 
out of the lubricant. 

8. Same as in the case of dry 
surfaces. 



The well-known expression for the friction of sliding is 

r = /xP. (136) 

Where F is the resistance to sliding, or the friction, P is the total 
pressure between the surfaces, and //, is a constant called the coeffi- 
cient of friction. This same expression will hold for the frictional 
resistance of lubricated bearings, but in this case /x is not the 
coefficient of friction of dry sliding, but is a constant depending 
upon the velocity of rubbing and the intensity of pressure between 
the surfaces. 

104. The Nominal Area of Bearing. — It will be shown further 
on how the actual pressure on a journal varies from point to point, 
being a maximum at the crown and least at the two sides. For 
practical calculations, however, the pressure is assumed to be evenly 
distributed over the projected area of the bearing. Thus, in Fig. 
102, d=: diameter of bearing; 1 = length; and the projected area is 
dxl. If P = the total load on the bearing, then the pressure per 
square inch of projected area is 

P 

13 



(137) 



190 ENGINEERING MECHANICS. 

105. TJie Worh Expended in Friction. — If a shaft or journal 
makes 'N revolutions per minute^ its surface velocity is 

12X60 ^^^^ per second. . (138) 

(12 inches = 1 foot; 60 seconds = 1 minute; d in inches.) 
From equation (136) Ave have resistance, or force, F = /aP, but 

Fxdist. = work; hence work of f riction = F X dist. = F X 
foot-pounds per second, or 

Work = /xP ^^Yac) foot-pounds per second, (139) 



TrdN 

12X60 



Fig. 102. — Area of Bearing. 

from which the work lost in friction can be found, provided the 
value of fx is known. 

106. The Value of /x for Journals. — This value has been deter- 
mined by numerous elaborate and carefully conducted experiments, 
the most generally accepted being those by Mr. Beauchamp Tower, 
with oil-bath lubrication (that is to say, with practically perfect 
lubrication) from which it is found that 

^ = c^, (140) 

where c = a constant; v = velocity of journal surface in feet per 
second; p = pressure per square inch of projected area of journal in 
pounds. 

In a general way the experiments consisted of a very great num- 
ber of trials at varying speeds and pressure, during which the fric- 



FRICTION xVND LUBRICATION OF BEARINGS. 



191 



tion was measured, and the value of /x calculated from the relation 
given by equation (136) ; /.i, v and p being thus known, the value of 
c is calculated from equation (140). 

The mean values of c thus obtained for different lubricants are 



given as follows : 


















TABLE 18. 








IiUbricant 


) 


Olive 
oil. 


Lard 
oil. 


Sperm 
oil. 


Mineral 
oil. 


Mineral 




1 


grease. 


Mean value of c 




.289 


.281 


.194 


.276 


.431 



During these experiments the pressures at which seizing occurs 
with the different lubricants were also determined, and were found 
to vary from about 300 to 600 pounds per square inch, according to 
the viscosity of the lubricant. It is from such experiments as these, 
modified by the results of practical experience, that tables of limit- 
ing bearing pressures, such as that given in Art. 99, Table 17, are 
obtained. 

It must be understood that the above values of c are for practi- 
cally perfect lubrication, such as oil bath, or forced systems. When 
the lubrication is not so perfect the friction is much higher and is 
generally very erratic. 

From the above, then, we have the frictional resistance with 
perfect lubrication, 



IX 



but 



or 



hence 



P 



dXl 



P. 



— ==dxl; 
P 



F^cxdxlx Vv, 



(141) 

where d = diameter of journal; 1 = length of journal, both in inches; 
v = velocity of journal surface in feet per second; and c = a constant 
depending on the lubricant, values of which are given in Table 18. 
For a more detailed discussion of this subject see Unwin, " Ele- 
ments of Machine Design.^^ 



192 



EXGINEERI^^G MECHANICS. 



107. The Distribution of Presswe in a Bearing. — In Beauclianip 

Tower^s experiments, oil-bath lu- 
brication was used in order to 
insure a regular and nniform sup- 
pi}^ of oil. It was found that the 
condition of the journal and brass 
was, as shown, greatly exaggerated 
by Fig. 103. The oil is carried in 
at the " on " side between the 
journal and the brass and forms 
a continuous film, being carried 
out again at the ^^ off " side. The 

brass is completely oil-borne by this film, which supports the whole 

load on the bearing. 

The film is thinnest, and the pressure greatest, along a line 

parallel to the axis of the journal, but on the " off " side of the 

center line at the top. 




Fig. 103.— Oil Film. 




Fig. 104. 



Fig. 104 shows the distribution of the film pressure, the pressure 
being measured at the several points indicated. N"o attempt is made 
to draw Fig. 104 to scale, as it is intended only to show in a general 
way the relative pressures at the different points. 

It will be seen that it is impossible to feed oil into a bearing 
carrying a load constant in direction by means of a hole in the top 
of the cap brass, as this will lead the oil close to the point of greatest 
pressure. In fact, in such a case the oil instead of entering, will be 



194 ENGINEERING MECHANICS. 

For naval engines a rather hard white metal is used, but care 
must be taken that the metal is not so hard as to be brittle at thq 
same time. 

The relative advantages and disadvantages of white metal linings 
for bearings are summarized as follows by Dr. Goodman : 

SOFT "WHITE METALS FOR BEARINGS. 
Advantages. Disadvantages. 

The friction is much lower than Will not stand the hammering 

with hard bronzes, cast iron, etc., action that some shafts are subject 
hence is less liable to heat. to. 

The wear is very small indeed The wear is very rapid at first 

after the bearing has once got if the shaft is at all rough; the 
well bedded (see disadvantages). action resembles that of a new file 

on lead. At first the file cuts 
rapidly, but it soon clogs, and then 
ceases to act as a file. 

It rarely scores the shaft, even it is liable to melt out if the 

if the bearing heats. bearing runs hot. 

It absorbs any grit that may get If made of unsuitable material 

into the bearing, instead of allow- it is liable to corrode, 
ing it to churn round and round, 
and so cause damage. 

110. Thrust Bearing. — Consider first a flat pivot as illustrated 
by Fig. 105. 

The total thnist P is assumed to be evenly distributed over the 
whole surface of the end of the shaft; then the pressure in pounds 
per square inch is 

P 



rR^ 



(142) 



The pressure on an elementary ring whose radius is r and width 
dr, is 

27rrdrp; (143) 

the friction of this ring is 

27rrdrpX)U,, (144) 

and the work of friction, if N = revolutions per minute is 

27rrdrp^ X 27rrN inch-pounds. ( 145 ) 



FRICTION AND LUBRICATION OF BEARINGS. 

Tlie work of friction over the whole surface is 

47r-p/AN rMr=r ^ . 

Jo o 

Substituting the value of p from equation (1-1:2) gives 

OTTXl" 

or the work in foot-pounds per minute is 

27r /.NPD ^ ^PDIN^ 

3x13 5.73 

or the horse-power absorbed by the pivot friction is 

-p,p^ /xPm^ 

* • 189,000 * 



195 



(146) 



(147) 



(148) 





Pig. 105. 

When the thrust cannot be taken on the end of the shaft, as is the 
case with the propeller thrust in marine engines, a number of thrust 
collars are provided as in the familiar thrust bearing. In some 
cases, where the thrust is not gTeat, a single collar is used, but 
where the load is large, the single collar necessitates a large diam- 
eter in order to keep the thrust pressure within allowable limits. 
This increases the loss due to friction on account of the increased 
velocity of rubbing, and the difference of velocity at the outer diam- 
eter of the collar and that at the surface of the shaft causes unequal 
wear. In practice, therefore, the outer diameter of the collars is 
made not more than one and a half times the diameter of the shaft. 
The necessary area of thrust surface is obtained by using a number 
of collars. 



196 



ENGINEERIXG MECIIAXICS. 



In naval engines the tlirnst bearings are made of the horse-shoe 
type (see Barton^s "Naval Machinery'^), and each ring can be 
independently adjusted. Each ring has its own water circulation 
and oil feed, and the lower parts of the thrnst collars on the shaft 
dip into a bath of oil, which is also cooled by circulating water. The 
horse shoes are faced with white metal. 

The work of friction in the case of the collar-thrust bearing is 
obtained as follows : 

Starting with equation (146) above, the limits of integration are 
between E^ and E2, see (Fig. 106). 



P 



y/^y 
/y//^A 



"^ 



f 




Fig. 106. — Thrust Collar. 

The expression for the work of friction thus becomes 
47r2p/.N r'r=^dr=:f7r^p/xI^(Ei^-E/) ; 

JR2 

also equation (142) takes the form 

_ P . 

P~7r(E,2_E22) ' 

so for this case we have the horse-power absorbed 
H.P 



(149) 
(150) 



(151) 



/xPN(E,^-E/) 

94538(Ei2-Eo-) * 
The lubrication of such bearings is always very good, being bath 
lubrication in character, so the coefficient of friction is rather low, 
the value of ^u, being from .007 to .01, 

/x=.01 

being a value commonly used in practice. 



rmCTIOX AND LUBRICATION OF BEARINGS. 197 

111. Design of Thrust Collars. — Example: Design the thrust 
journal for a steamer from the following data, and find the power 
lost in friction of the thrust bearing. 

Speed knots 22 

r. H. P. (each engine) 11500 

Diameter of shaft 17%" 

Outside diameter of collars 27" 

R. P. M 120 

Combined efficiency of engine, shaft and propellers 65%, say % 

It is evident tliat the thrust multiplied by the distance the ship 
advances in a minute must equal the work delivered to the thrust 
block per minute. This work, in foot-pounds per minute, 
= 1 XL H. P. X 33,000. 

Let P=: total forward thrust (each engine). 
S = speed in knots per hour, 
then 

^ — — — = speed in feet per minute. 

PX 8x^080 
60 
or 

P- LH. P.x33,000x2x6 _ 11,500x33,000x2x60 
~ 3 X 6080 X S ' ~ 3 X 6080 X 22 

= 113,500 (by omnimeter). 
The area of each collar is 

I (Di^-D/) = ^ (272-17:752) =-^ (729-315) -^^^i^i =325.8. 

The allowable pressure per square inch of thrust collar is 60 
pounds (see Table 17) and the total pressure on each collar is, 
therefore, 325.8x60 = 19,548 pounds. The number of collars is 
then 

113,500 ^Q . 

-19;548=^-^^^^^^^^^- 

But in the horse-shoe type of thrust bearing the horse shoes encircle 

only about one-half of the shaft, and, therefore, we must double the 

above number, giving 12. 

Use 12 collars. 
In naval practice the thickness of each collar is 

D,-D, 
4 



fxLH.P.x33,000 = 



19<S EXGIXEERING MECHANICS. 

This is much greater than is required to resist the shearing force, 
but is used to allow for the possibility of the whole thrust coming 
on a few of the collars due to improper adjustment, and also to 
provide a sufficient space between adjacent horse shoes for the inde- 
pendent adjusting nuts. 

From equation (151) the horse-power lost in friction on one thrust 
block is 

TT p ^ jttPF / E,^-E/ \ _ . 01x113,500x120 /13^5^-^8^3 
• • 189,000 VEi^-E.V" 189,000 ll3:5--8::8?2 

_ .01x113,500x120x1 761 ^9, ,. 
94,538x103.8 

QCESTIONS AND PROBLEMS. 

Explain in general terms the effect of friction in bearings and 
how it is reduced and controlled. What is meant by " working 
heat '^ of a bearing ? Explain the " seizing '^ of a bearing and how 
it may be prevented. 

Explain the difference in the coefficient of friction for dry sliding 
and the corresponding constant for lubricated surfaces. What is 
the nominal area of bearing? Deduce an expression for the work 
of friction of a bearing, in terms of ^, load, diameter of journal and 
E. P. M. State the relation between /x, the velocity of rubbing and 
the intensity of pressure of a bearing. 

Make a sketch showing the action of the oil between a journal 
and its brasses when lubrication is ample. Show also by a sketch 
the distribution of the oil pressure in a loaded bearing, both trans- 
versely and longitudinally. Where should the oil enter a bearing 
carrying a steady load ? How and why does this differ from a crank- 
pin bearing ? 

Describe the naval system of forced lubrication. 

Discuss the use of anti-friction metal. 

Problems. 
1. The diameter of the cylinder of an engine is 30.7 inches; 
mean effective pressure = 40 pounds per square inch; E. P. M.=:140; 
diameter of crank pin = 10"; length of crank pin = 10". Perfect 
lubrication^ using olive oil, find 

(a) Eubbing speed per second. 

(b) Pressure per square inch of bearing surface on pin. 

(c) Value of /x. 



FRICTIOX AND LUBRICATION OF BEARINGS. 199 

2. An engine has a cylinder whose diameter = 30 '.'7; mean effect- 
ive pressure = 40 pounds per square inch; E. P. M = 140; stroke 
= 24''; diameter of crank pin = 10"; length of crank pin = 10". Lard 
oil is used with a system giving perfect lubrication. Find fi. Find 
the percentage of the total work lost in crank-pin friction. (Sug- 
gestion: Total energy in foot-pounds per minute = 2pLAN.) 

3. An engine, whose cylinder diameter = 38 J"; stroke = 48"; 
diameter of crank pin = 19"; length of crank pin = 22J"; M. E. P. 
= 118 pounds per square inch; E. P.M. = 128, uses mineral oil 
with flooded lubrication. Find the value of /x; and the percentage 
of the total energy lost in crank-pin friction. 

4. Find the number and thickness of the thrust collars and the 
horse-power lost in friction of the thrust bearing of a naval engine 
from the following data: 

Speed in knots Igi^ 

I. H. P. (each engine) 8250 

R. P. M 125 

Diameter of thrust shaft — inches 15i/^ 



CHAPTER XV. 

Notes on the Design of Crank Shaft, Cross-head Pins, fj'c* 
practical problem v. 

112. Cranl' Shaft. — In calculating the dimensions of a crank 
shaft for strength, the theoretical aggregate mean effective pressure 
is found and the actual pressure expected is determined by compari- 
son with the results given by the trials of engines of the same type, 
pressure and rate of expansion. In other words, a value for the 
efficiency of the valve gear, or the percentage of the actual to the 
theoretical mean pressure must be obtained (Seaton). 

This efficiency is obtained as follows: Let r= total rate of 

area of L. P. cylinde r 

^ ~ area of H. P. cylinder x cut-off in H. P. cylinder * 

Then the total mean pressure P^ is found from 



^^1 + log^ (152) 



OT, by using table, Barton, p. 522. 

P, = P^-p,. (153) 

This calculated value of Pe is compared with the actual value 
found from the trial trip, and the efficiency is found from 

_ actu al p 
calculated 

The above method may be used when the I. H. P. is not pre- 
viously known, but in all practical cases the I. H. P. is the first 
thing determined, and all detail calculations are based on it. 

113. Bearings. — Pressure per Square Inch of Projected Area with 
Regard to Heating: Seaton. — The bearing surface of crank pins 
must be such that the pressure per square inch does not exceed 500 
pounds, and, in the case of merchant ships where room will permit 
of larger pins, 400 pounds should not be exceeded. When the brass 
is recessed so that it bears only on parts of the shaft, the actual bear- 
ing surface should not be exposed to more than 600 pounds per 
square inch. Main hearings of screw engines, when room admits, 
* From " Notes on Machine Design," revised and enlarged. 



NOTES ON DESIGN OF CRANK SHAFT, CROSS-HEAP PINS, ETC. 201 

should be of such a size that the pressure does not exceed 200 
pounds, measuring the whole of the bearing, or 300 pounds on the 
actual bearing surface. 

It is found that the channel ways take up about 20;^ of the 
bearing surface, leaving about 80fc as the actual bearing surface. 

The length of crank pins is from 1 to IJ times the diameter of 
the pins. Vertical engines usually have space for a length of 
bearing equal to IJ diameters. The length of each main journal 
is from 1 to 1^ times the diameter of the shaft. Vertical engines 
have usually a length of bearing equal to 1^ diameters. 

Foley's Engineers' Reference Book. — '' No hard and fast rule can 
be laid down. The engineer must be guided by circumstances 
and experience. The velocity with which the surfaces move on 
each other must be considered, and also whether the direction of the 
pressure alternates. In the case of a cross-head pin, the motion 
is not only slow, but the pressure alternates as well; hence the high 
pressure permissible. The following may be a rough guide : 

" Cross-head pins, 750 pounds ; not to exceed 1000 pounds. Crank 
pins, 300 pounds; not to exceed 500 pounds. Main bearings, 250 
pounds ; not to exceed 500 pounds.^' 

The Bureau of Steam Engineering uses the following formula 
as guides : 

For crank-shaft journals: pVv<8000(l). Where p = pressure 
per square inch and must not exceed 500 pounds; v = velocity of 
surface in feet per minute. 

For crank pins: pVv<15, 000(2) and p must not exceed 700 
pounds. 

114. HoUoiv Shafts and Pins. — ^^The diameter of the equivalent 

hollow shaft may be calculated by the equation d^ = ^ ~ ^ - . 

(See Art. 95.) ' 

Foley^s practice is : 

When dia. of hole = .4 of d add 1^ to dia. of solid shaft. 



iC 


= .5 


iC 


2 


a 


(( 


=:.6 


a 


5 


« 


C( 


==.7 


cc 


10 


a 



This brings very nearly the same result as solving the equation 
above, which must be done by assuming some ratio between the 
diameters of shaft and bore hole or by assuming one diameter. 



202 



ENGINEERING MECHANICS. 



For convenience of interpolation the above table from Foley has 
been plotted as a chart, which is used as follows : Having first 
found the diameter of the solid shaft necessary and having fixed on 
the diameter of the hole through the axis of the shaft, find the 
decimal fraction from the ratio diameter of hole to diameter of 
solid shaft, this decimal is shown at the left hand edge of the chart; 
ran across horizontally to the curve; drop down vertically to the 
bottom edge of the chart, where will be found the percentage to be 
added to the diameter of the solid shaft to give the diameter of the 
equivalent hollow shaft. 

Graphic approximate solution of d = a/ -^^ — ~ to be used 
stead of Foley's practice : 



m- 



.7^- 






Si 



^^ 



2. 



\ 



^ 



!V- 



.J^ 



/Z 27o 57o ^X SZ 6Z 77o SZ 3% /OZ 
/^erCe/it fo ^ ac/c/ed ^o Solid S/za/6. 



Fig. 107. 



The Bureau of Steam Engineering makes the diameter of the 
crank pins one inch larger than that of the shaft journals. The 
length of the pins is calculated so that a certain pressure, about 
400 pounds per square inch, is not exceeded; this length being 
generally \\ times the diameters of the pins. 



NOTES ON DESTGX OF CRAXK SHAFT, CROSS-HEAD PINS, ETC. 203 

115. Calculations. — The engine for which the following calcula- 
tions are made is that of the U. S. S. " Ealeigh." 

Data: Vertical, inverted cylinder, direct-acting, triple-expan- 
sion, twin-screw engines. Cylinder diameters, 36 inches, 53 inches, 
80f inches. Stroke, 33 inches. Eevolutions per minnte, 16-1:. 
Boiler pressure 160 pounds, by gauge. First receiver pressure, abso- 
lute, 60 pounds (assumed). Vacuum, 24 inches. Cut-off in H. P. 
cylinder, .7 stroke, I. H. P., 9863.2. 

M^^f;^_-1^- r=l.5; dist. between bearings, 53". f = 9000. 
Mean ^ 

116. Cranh Shaft. — For one set of engines 

I. H. P.=r4931.6. 

I. H.Px 33,000x12 



Mean T. M. of after enoine 

T TT P 

X 63,000. 



27r X Rev. 
LH.JP. 
Rev." 



TV/T • rr ^,T 4931.6X63,000X1.5 o o^n nnn i 

Maximum T. M.=: ^ =2,840,000 and 



Max.T. M.2 = T- = 8,075,000,000,000 (by omnimeter). 

Wl 

M, the bending moment, = ^ , where W = maximum turning 

o 

. Max. T. M. 2,840,000 -.rvoonn i i ^- + u 

torce= = = ' ^^' — —172,200 and 1 = distance be- 

crank 16.5 

tween bearings = 53". 

.-. M= ^ = _12M^X53_ =1,140,000, and M^ = 1,302,000,- 



000,000. Te = equivalent T. M. = M-}- VM^ + T- = 4,202,000. 
T, = f% = f ''^ .'. Dia. of shaft = d= \/ ^^^j^ 

- y—T- = V — 9000 ^^•^^- 

This is for a solid shaft. The shaft in question is to have a 
6-inch axial hole. Referring to Fole3^'s practice, as above : 

Dia. of hole= -.q o~ ^ = 45;^ of dia., nearl}^ By interpolating 

in the table or by the use of the chart we get 1.5^ increase for 
d2 = .45 di. Then 13.35 x 1.016 = 13':56, or dia. of shaft, by calcu- 
lation for strength, = 13 yV'- 



304 EJs^GINEERING MECHAXICS. 

Length of Main Journals. — In order to obtain this, we must find 
first the thrust E on the connecting rod. This is calculated for the 
H. P. cylinder. 



R R 

Effective load on H. P. piston = P = areax (p^ — pr, or initial 
abs. press.-rec. press.) =1018x (175-60) =117,100. Then E = P 
sec(sin-ii) =121,000 pounds. 

Then — > the thrust on each bearing, = 1^'— = 60,500 pounds. 

From notes on " Bearings/^ allowing a pressure of 190 pounds per 
square inch of projected area. 

The projected area required=^5^ 3=318.1 = lxd. 
But d = 13yV whence 1= ^^^-^ =23.45. 

1 23 45 

From this,— j- = ' =1.73, which is too large, as there is 

seldom fore and aft length allowed for such a coefficient. Therefore, 
the dimensions obtained must be changed as follows : 

l = 1.5d .'. lxd = 1.5d^ = projected area. 

.*. 1.5d2 = 318.1 and d = 14.57, or MyV'- 

.-. l = 14iVx 1.5 = 21.82, or 2111''. 

The projected area is, then, 14Y^g-x21|f= 316.5 square inches. 

The pressure per square inch in this area= P '—^ =191 

316.5 

pounds, which is within the limit of 200 pounds. The area of the 

actual bearing surf ace = 316. 5 X. 8 = 253. 5 square inches, and the 

pressure per square inch of this area= ' - = 238.4 pounds, which 

is within the limit of 300 pounds. 
The Bureau of Steam Engineering's check for speed: velocity 
7rdxl64 



12 



where d = diam. main bearing in inches, 164 is 



revolutions per minute = 'L><l_^i><_A_^i ^626. .*. Vv = 25.02 ft. 



NOTES ON DESIGN OF CRANK SHAFT, CROSS-HEAD PINS, ETC. 205 



14 



206 



ENGIXEEKTNG MECHANICS. 



p, the pressure per square inch of projected area (from above), 
= 191 pounds. Then pVv = 191 X 25.02 = 4780, which is within the 
limit of 7500. 

For the dimensions of the main bearings^ we have, then : 

Diameter = 14yV'; length = 21if"- 
117. Cranh Pins. — All crank pins are made of the same dimen- 
sions as the after pin, which is calculated for bending moment. To 

. I determine the bending moment on the 

-S-] r&-J t-S] |-S- after pin, consider the turning force of 
' — ' ' — ' the H. P. and I. P. engines to be trans- 

mitted directly back to the forward end of the L. P. pin; or, what 
is the same thing, consider two-thirds of the turning force of the 



/ ^/ 



/ 7m ox. ^^ f w 



orM 



^Cm^^""'^ 



f/r. 




/a 



'/Tfoac. 






5 ^^r^^/r^^J^ 



Fig. 108. 



or^K 



three engines transmitted to this point. Then, the pressure or load 
transmitted to the forward end of the L. P. pin is equal to two- 
thirds of the maximum twisting moment, divided by the length of 

the crank. Let ^^^^ "^I ^- or -^^ =K and length of pin = l. 
crank crank ° ^ 

Then, K=: --—^^V — = 172,200. Since the transmitted force 
lb. 5 

acts on the forward end of the pin, its bending moment = |K1. 
In addition to this load on tlie forward end, the pin takes the bend- 

K 



XI 



ing moment of its own engine, which is 



, since the pin is 



K 



fixed at the after end and carries a load -7- uniformly distributed. 



^OTES ON DESIGN OF CRANK SHAFT, CROSS-HEAD PINS, ETC. 207 

That is to say, the turning effort of the H. P. and I. P. engines 
carries around the forward web of the L. P. crank, so that it offers 
no support to that end of the pin in resisting bending. The resist- 
ance of the propeller tends to prevent the free swing of the after 
web, so in effect the whole bending moment on the L. P. pin tends 
to break it off at its junction with the after web. 
Hence the total bending moment on the pin is 

M = 2^ +^X ^ = «KI.* (154) 

Substituting this bending moment in the formula for a beam subject 
to bending gives 

From notes on " Bearings ^^ --r- is from 1 to 1.5. Take it as 1.25, 

since naval engines seldom have enough, fore and aft space for the 
ratio 1.5. Then 

d= ,/ 5X 172,200x1.25x10.2 _ p. -j.^.. 
V 6x9000 -i^.'^o-i^4 • 

Then, 1 =:1.25d = 1.25 x 14.25 = 17.80, or 17||^ 
These dimensions are for solid pins. For hollow pins apply the 
table given above. 

Dia. of hole = --^^ ^ = 42^ of d and 14.25 X 1.02 (interpo- 
lating in table or using chart) =14.55, or 14^", = dia. of hollow pin. 
Also, length of hollow pin = 14325-'' x 1.25 = 18.2, or 18 3^". 
These are the dimensions of the pins for strength. 

* In tlie case of the four-cj^linder triple expansion engine with two 
L. P. cylinders, the bending moment on the after crank pin (v/hich is 
always a L. P. pin) becomes — 

(a) When the two L. P. engines together develop % of the total power 
of the engine 

M = iKl + |Kl+ i^ =||K1 (155) 

(b) When each of the four cylinders develop 14 of the total power 

1 Kl 
Mzzz3KI+ i^ =7K1 (156) 



^08 ENGINEERING MECHANICS. 

Chech for Heating. — Pressure per square inch of projected area 
must not exceed 500 pounds. Assume a pressure of 450 pounds 
per square inch, and we have : 

Projected area of pi„^I°ad or thrust E on pin ^ 121,000 ^^ggg 

450 450 

square inches. 

Eeferring to the calculated dimensions for the crank pins, 14^^'' 
Xl8^^ the projected area=265 square inches. This area, there- 
fore, though ample for strength, is too small to prevent heating, for 
which an area of 268.8 square inches is required. Therefore, the di- 
mensions of the pin must be increased : 

Then (dxl.25) d = 268.8. .*. d=14r67, or 14fV', and 1= 
1.25d=18r3, or 18yV'; and the projected area = 14|i"xl8A''= 
268.5 square inches. The pressure per square inch on this area 

121,000 ,^. -, 

= ^68X =^^^ P'^'^^'- 

The actual projected area of bearing surface is, then, 268. 5 x. 8 
= 214.8 square inches; and the pressure per square inch on this 

area=— ^-^-^— =563 pounds, which is inside the limit of 600 

pounds. 

Bureau of Steam Engineering's check for speed: From notes 
on "Bearings,'' equation (2) : 

pVv< 15,000, p not to exceed 700 pounds. 



Vy 



>Xdxl64 ^ / 3.14xl4|ixl64 ^^^ ^ f^,^ 

V 12 V 12 



the slight difference in v, owing to the angularity of the connect- 
ing rod, being left out of the calculation. 

Also, from above, p = 563 pounds; then, pVv = 563 X 25.1 = 
14,300, which is within the limit of 15,000. 

It is seen, then, that the dimensions found fulfill all conditions, 
the changes in diameter keeping the hollow pin amply large for 
strength. 

The dimensions of the crank pins are, then: Diameter = 14|-^''; 

length = 18yV'- 

Bureau of Steam Engineering's practice: 

Dia. of shaft = 13yV'- Then dia. of pins=i^TV'- Length of 
pins = 1.25 X 14yV' = 18.2, or 18yV'- Bearing pressure per square in. 



NOTES ON DESIGN OE CRANK SHAFT, CROSS-HEAD PINS, ETC. 209 
i? • ^J 121,000 121,000 .K^o ;i 1 • -u 

01 projected area=: . ' — = 9 v'i q~3~ ~^^"-^ pounds, which 

1 X d 1 4y-g- X 1 O Y g- 

is within the limit of 500 pounds per square inch. 
These dimensions are slightly smaller than those 
found by the above calculations, so the dimensions .,. 
found in " check for heating " will be taken, = 

d = 14|i"andl = 18yV'- 

117a. Aftei- Arm of After Crank. — Since the lev- c- 
erage is at the pin, the greatest thickness is required 
at Cd. Fig. 109. 

The bending moment at cd is the entire twisting moment on the 

bh^ 
crank shaft at the after bearing, and —^ = modulus of section 

(rectangular) . Therefore 

^'f = Max. T. M. (158) 



. , /Max. T. M. X 6 , , Max. T. M. x 6 

.•.h=^ b>?f ^^db= ^.^^j . 

Bureau of Steam Engineering's Method : The check used by 
the Bureau is to make the width of the crank arm across the face, 
h, about two inches more than the diameter of the shaft, and then 
compute the value of the thickness b^ by the formula above for b, 
or by using a formula given by Unwin, 

bh2 = cd^ (159) 

giving c a value of .9 for steel. 

1st Method: Dia. of shaft^lSyV'- •*• h = 15yV', and 
._ 2,840,000x6 ^y^-, 
(15tV)'x9000 • 

2d Method: bh2 = cd^ .*. b = ^^^^^?|^- = 9r27. 

The average of these two methods gives 8.54 inches. The dimen- 
sions of the after arm of the after crank are, then, b = 8U''; 

If, in working out the design of the crank shaft as a whole, the 
sum obtained by adding the length of crank pin plus twice the 
thickness of the web is greater than the distance assumed between 
main bearings, then either this distance must be increased, or the 



210 ENGINEERING MECHANICS. 

length of the crank pin decreased. In either case the calculations 
must be made a second time with the new conditions. 

118. Cross-head Journals. — Assume that two-thirds of the thrust 
E of the connecting rod may come on each journal. This allows for 
inaccurate adjustment and irregular wear. 

fR = P = f Xl21,000 = 80,666 pounds. 
¥1 . (d,^-d,^) 

2 ^ 32 di "• 

In this case, let -^ =1.1, and the axial hole d2 = 4''; f = 9000 for 

steel. The ratio of -r is taken small to prevent too great a spread 

of the jaws of the connecting rod, thus making a lighter rod. 
Substituting, 

i^><8Mm^ =9ooox vx,vx ii^zdin . 

2 7 32 ^^ 

Solving for d^ we have d^ = 7" 4.2, or 7if ". 
Then, l = l.ld = l.lx 7i|" = 8.15, or 83V'. 

This gives dimensions for strength. The projected area of the 
bearing surface = 7^f"x8-3V' = 60.4 square inches and the pressure 

per square inch of projected area= ' =:1335 pounds. 

Check for Heating. — From notes on " Bearings " we find the 
pressure per square inch of projected area allowed for cross-head 
pins to be 750 pounds and not to exceed 1000 pounds. This shows 
that the pins are too small to prevent heating. 

Assume a pressure of 750 pounds per square inch. Then the pro- 
jected area required = = ~«W~ =107.6 square inches. 

Since l = l.ld, l.ld2=:107.6 and d = 9:'89, or 9J". .M = l.ld 
= 10':85, or lOff . 
The projected area is, then, 9f''x 10||-'' = 107 square inches, and 

the pressure per square inch = — ' =753 pounds, which is within 
the limit of 1000 pounds. 

The dimensions of the cross-head pins are, then: Dia. = 9J"; 
length = 10| J". 



NOTES ON DESIGN OF CRANK SHAFT, CIIOSS-IIEAD PINS, ETC. 211 

119. Shaft Couplings. — The couplings are forged solid with the 
shafts, and are subjected to a 
twisting strain tending to shear 
the coupling off leaving the shaft 
cylindrical to the ends, so that 
the coupling can rotate on it like 
a washer on a bolt. Let t = the 
thickness of coupling at a radius 
R from the center of the shaft. 




Then 27rRt is the area of the sec- Fig. 110. 

tion and 27rEtf its resistance to 

twisting. Its moment around the center of the shaft is 27rR^tf, and 

this should equal the twisting moment on the shaft, which is 

7rfd^-^16. Equaling these, we have: t= ^-^2 . From this it 

will be seen that, as R increases, t decreases. When R= „ , or where 

d^ d 
the coupling joins the shaft, we have, t= ^-^ = .r . This gives 

the tliickness at this point for strength only, but is found to be too 
small for stiffness, and allows nothing for the loss of material due 
to the coupling-bolt holes. Seaton states that, from practical con- 
siderations, the thickness of the couplings should not be less than 
the diameter of the coupling bolts. He also states that, to allow for 
the decrease in strength due to cutting holes for the bolts, the thick- 
ness should not be less than .3 X dia. of a shaft subjected to twisting 
only. This is the same as given in Art. 101, Fig. 101, where the 
usual dimensions of couplings are given and are the ones to be used 
in the design. 

120. Coupling Bolts. — The number of bolts depends, to a certain 
extent, on circumstances. If it be desirable to keep the outside 
diameter of the coupling as small as possible, the number of holes 
should be increased; this decreases their diameter, and brings their 
centers nearer to the center of the shaft, thus reducing the outside 
diameter of the coupling. In general, when engines have two cranks 
with the shafts in duplicate pieces, there should be an even number 
of bolts. When there are three cranks and three interchangeable 
parts of the crank shaft, the number of bolts should be a multiple 
of three. In this case six is the number usually taken, and this 
applies to the couplings of the line shaft also. 



212 E>s"GI^TEEEING MECHANICS. 

Diameters of Coupling Bolts. — Let K = radius of bolt circle, 
which is about .80 of the diameter of the shaft. The coupling bolts, 
if well fitted, will be subjected to shearing only. 

Let fs = shearing strength of material = f to f of f^ 

P X K = T. M. = ^ f fd^ and, at bolt circles, the force P is -^~- . 
Let 8 = dia. of bolts, and n = number of bolts, then: 

In the case under consideration, n = 6, K = 10.85 and d = 13-3^. 



8 = 3.57, or 3 



Questions and Problems. 

Describe clearly the method of designing a crank shaft (includ- 
ing crank pins and webs), showing to what stress the various parts 
are subjected. After finding the sizes of journals and pins for 
strength, what further check is applied ? 

Problems. 

1. Find the dimensions of the main journals of a vertical triple- 
expansion three-cylinder engine; twin screws; I. H. P. = 10,000; 
cylinder diameters, 33^ inches, 51 inches, 78 inches; stroke, 48 
inches; initial pressure, 180 pounds per gauge; vacuum 26 inches; 
cut-off in H. P. cylinder, -^^ stroke; 1st receiver pressure 70 

Max T M 
pounds (absolute); ' ' -- =1.5; distance between bearings, 

4 feet 5^ inches; f^ = 9500; shaft to have 7-i-inch axial hole; 
E. P. M. = 116. 

2. With the same data as problem 1, find the dimensions of the 
crank pins; axial hole through pins = 6''; ratio 1 to d=1.25. 

3. Find the dimensions of the after crank area of a three-cylinder 
triple-expansion engine from the following data : diameter of crank 
shaft=14f"; I. H. P. (one engine) = 5000 ; E. P. M = 116; ff = 
9500. 

4. Find the dimensions of the crank shaft couplings and bolts 
for a three-cylinder triple-expansion engine; I. H. P. (one engine) 
= 5000; E. P. M. = 116; fi = 9500; diameter of sliaft=14f^ with 
7J-inch axial hole; shaft in three interchangeable sections; for 
bolts take fs = Jf^ 



NOTES ON DESIGN OF CRANK SHAFT, CROSS-HEAD PINS, ETC. 213 

121. Sequence of Calculations of the Crank-shaft ProUem: 

1. Find the mean T. M. on shaft abaft the after engine. 

2. Find the maximum T. M. 

3. Find the bending moment, M. 

4. Find the equivalent T. M., Tc=M + VM^ + T^^a^. 

5. Find the diameter of the solid shaft. 

6. Find the diameter of the equivalent hollow shaft either from 
the known size of the hole, if given, otherwise by assuming B^^^^B-^. 

7. Find the length of the journals, pressure per square inch of 
projected area less than 200 pounds. 

8. Check for reduction of area due to oil grooves, less than 300 
pounds per square inch. 

9. Check for speed of rubbing and pressure px Vv<7500; giv- 
ing the final dimensions. 

10. Find K. 

11. Find bending moment at after end of after crank pin. 

12. Assume a value of the ratio -r and solve for the diameter of 

d 

crank pin. 

13. Eeduce diameter to diameter of hollow pin, and find length 
of pin. 

14. Check for pressure, not to exceed 500 pounds per square inch. 

15. Check again for reduction of area due to oil grooves, not to 
exceed 600 pounds per square inch. 

16. Check for speed of rubbing and pressure, pxVv< 15,000, 
giving final dimensions. 

17. Note whether or not the dimensions of crank pin, as found 
above, are smaller than the Bureau practice of making diameter of 
crank pin 1 inch larger than diameter of shaft. If it has worked 
out smaller, then make diameter of crank pin = diameter of shaft 
+ 1" ; if it has worked out larger take largest dimensions. 

18. Find dimensions of crank webs, making h = diameter of shaft 
+ 2". Using both the bending method and Unwinds method, and 
taking the mean of results for b. 

19. Find thickness of shaft couplings. 

20. Select the proper number of coupling bolts, and find their 
diameter. 

21. Check diameter of coupling to see that the nuts of coupling 
bolts do not project beyond the circle of the coupling. 



214 



ENGINEERING MECHANICS. 



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NOTES ON" DESIGN OF CRANK SHAFT, CROSS-HEAD PINS, ETC. 215 









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CHAPTER XVI. 
I. H. P. Required for Given Vessel. Cylinders. 



Valve Chests. Pistons and Rods. 
Exhaust Pipes.* 



Valves and 
Steam and 



123. The general method is as follows : 

Data. — Displacement; length; breadth (extreme) ; mean draught ; 
area of immersed midship section ; speed required. 

Problem. — To design the propelling machinery of a naval vessel 
with the following dimensions and data: 

Displacement 13,700 tons 

Length 502 feet 

Breadth (extreme) 69 feet 6 in. 

Mean draught 24 feet. 

Area of immersed midship sect'n 1596 sq. ft. 

Speed required 22 knots 

Indicated Horse-power. — This is determined by two methods : 
Kirk^s " Analysis " and Froude^s " Law of Comparisons.^^ 

KirFs analysis : From a comparison of the trial trips of several 
vessels of similar lines, it is found that, reduced to a general basis 
of ten knots per hour, from 5.0 to 5.5 I. H. P. are required for each 
100 square feet of wetted surface. 

Wetted surface : Kirk's Analysis — 




Fig. 111. 



13,700x35 = Vol. in cubic feet, 

13 ,700X3 5 _ 300 -AH 
1596 -^^^-^^1^ 



* From " Notes on Machine Design," rewritten. 



218 ENGINEERING MECHANICS. 

124. Indicated H. P. Ijy Froude's Law of Comparisons. — Froude^s 
law may be stated as follows : If two ships have the same coefficient 
of fineness, same water lines, etc., but their dimensions have a ratio 
n, then for speeds s-,, Sg, S3, etc., of one with resistances r^, y^, rg, 
etc., the corresponding speeds of the other will be SiVn^, S2Vn, 
SgVn, etc., with resistances n^r^, n^rg, n^rg, etc. 

In similar ships, the cube of the ratio of dimensions equals the 
ratio of displacements, hence n^= ratio of displacements and 
Vn= (ratio of displacements)*. 

Let d, r, s and p represent the displacement, resistance, speed 
and I. H. P respectively of a given vessel whose data are known, 
preferably from a progressive speed trial accurately conducted; or, 
the same quantities may be obtained experimentally from a model 
of the proposed ship. 

Let D, E, S and P represent the corresponding quantities for 
the vessel whose power is to be determined. 



ThenS = s^^^ and R = r^, also p = rs. 
^w, P==RS-rsx ^ X y^ =pX ^ X y 



I. H. P. REQUIRED FOR GIVEN VESSEL. 219 

This law applies only to exactly similar ships. It is based on two 
fundamental assumptions : 

(a) The resistances of two similar ships vary as their displace- 
ments, that is: 

r 

also 



D D 3 /B' 



R = r^. (162) 

(b) The corresponding speeds of the two ships are proportional 
to the square roots of their lineal dimensions, that is : 

S=rsVn = s^-^. (163) 

Power is rate of doing work : .*. P = RS, and p = rs. 

Applying this method to the engine under design and assigning 
as a model a ship of similar lines and fineness and whose trial was 
accurately conducted, we obtain the following : 

Model Ship. Proposed Ship. 

d = 6000. D=:rl3,700. 

s= 16.5. S= 22.0. 

p=:5820. P= ?. 

Substituting and solving for S and P we have 
/'B ,^ ,^/ Vl3,700 



6/"]T V 

y_^=16.5X ^ 



6000 
16.5x1.148 = 18.95 corresponding speed. 



P-cTV-d 



^ = 5820 X 2.283 X 1.148 = 15,250. 
d 



This is the power necessary for the proposed ship at the speed cor- 
responding to the speed of the model. 

But our new ship is to have a speed of 22 knots. The power 
must therefore be increased in proportion to the cube of the speed. 

/ 22 
Indicated horse-power by Froude^s law = 15,250 x i 

= 23,850. 
Indicated horse-power by Kirk's analysis = 23,600. 
Mean of the two methods= 23.725. 



220 EN^GINEERING MECHANICS. 

In designing for large powers use mean to nearest hundred, or 
23,700; in this case, indicated horse-power of one engine = 11,850. 

125. Engines. — The t}^pe of engine, number of engines and other 
data are found from experience and best practice, and for this vessel 
the following are decided upon : 

Twin screws. Triple-expansion, vertical, direct engines, with the 
low-pressure stage in two cylinders. Independent pumps. Boiler 
pressure, 265 pounds (p. g-)- Cut-off in H. P. cylinder .75 stroke. 
Eatio L. P. to H. P. cylinders about 8.5 (including clearances). 
Clearance H. P., 15^; LP., 14^; L. P., 12^. Stroke, 48 inches. 
Eevolutions per minute about 125. Vacuum, 26 inches. I. H. P. 
(each engine), 11,850. 

126. Diameters of the Cylinders. — From the above data, the 
actual ratio of expansion in the H. P. cylinder, taking clearance into 

account, is: - , - , „. =1.278. 
.loi-. /o 

The total ratio of expansion of the engine is : Eatio L. P. to 

H. P. X ratio of expansion in H. P. = 8.5 X 1.278 = 10.86. 

The total theoretical mean pressure is: P^ = p^x ^^^ 

= 280x ^^^\l^i ^^^ = 280 X. 312 = 87.36. Or the table in 



" Barton," p. 522, may be used to obtain the value of ^ 



r 



A vacuum of 26 inches corresponds to an absolute pressure of 2 
pounds. Assume a back pressure of 2 pounds. Then the theoretical 
mean effective pressure is 

Pe =87.36-2 = 85.36. 

From the data of many trial trips of various ships, the actual 
mean effective pressure is found to be from 50^ to 60^ of the 
theoretical. This percentage is commonly called the " card effi- 
ciency." If we assume a card efficiency of 55^, the expected mean 
effective pressure is 85. 36 X. 55 = 46. 95. 

Using the well-known horse-power formula, 

PxLxAx2N _ 45.95x4x2xl25xA 

33,000 -J-L,«oU- 33,000 

we get 

11,850x33.000 Qooo • -k 

^^•^•= ^6.95X1000 =^''^ ^^"^^^ ''''^'' 
(for Ijoth cylinders) or 4161 square inches for each L. P. cylinder. 



I. H. P- KEQUIRED FOR GIVEN VESSEL. 221 

The area of the low pressure is thus calculated as if the whole 
expansion of the engine took place in this cylinder. The reason for 
this is fully explained in Barton, " Naval Engines and Machinery.'^ 
Area H.P. Cylinder. — Area H. P. cylinder (1 + H. P. clearance) 
area L. P. cylinder (1 + L. P. clearance) 
~ cylinder ratio ' 

or 

Azoto(both)(l + .12) 



Ah. p. (1 + .15) 



8.5 
8322x1.12 



. . i\7T P — —954 square inches. 

^•^•~ 8.5x1.15 ^ 

Area I. 'P. Cylinder. — The size of this cylinder has no effect on 
the total power of the engine and good practice shows considerable 
variation in its relative size, which is determined principally with 
the object of obtaining a proper distribution of work and considera- 
tions of balancing the engine as a whole. The following formula is 
frequently used and gives good results : 

Area I. P. Area L. P. 



1.0 to 1.1 V ratio of L. P. to H. P. areas 
or, using 1.0, 

8322 



1.0xV8.5 



= 2850 square inches. 



In determining the above areas the area of the piston rod has not 
been taken into consideration, therefore, the sizes of the cjdinders 
must be modified. 

Assume the size of the piston rod to be 8 J inches, which is about 
the size necessary for an engine of this size and power. 

One-half area corresponding to a diameter of 8^" = 28.37 square 
inches. Then, 

Total area H.P = 954 + 28.37= 982.37, 

Total area LP =2850 + 28.37 = 2878.37. 

Total area L. P. (each) =4161 + 28.37 = 4189.37. 

The corresponding calculated diameters are: 

H. P =35.4. 

I. P =60.6. 

L. P. (each) =73.1. 
15 



222 ENGIXEEEIXG MECHANICS. 

Then the diameters of cylinders used are : 

H. P =35-r] 

I. P =61" Iby 48" stroke. 

2 L. P. (each) =73" j 

Cylinder ratios ^~ = 8.46 ; ^^ =2.96 ; ^^ = 1.43. 

KoTE. — In case superheated steam is to be used the area of the 
H. P. cylinder must be increased in proportion to the increased 
specific volume of the superheated steam. Thus if 100° superheat 
is to be used, the H. P. area must be increased about 11^. 

127. Thickness of Cylinders and Various Parts of the Cylinder 
Castings. — The cylinders of modern engines are steam- jacketed, the 
jacket being formed by the space between the cylinder barrel and 
the liner. The Bureau of Steam Engineering uses the following 
formula : 

For cylinder liners t= ^ + C. (165) 

P = boiler pressure; d = diameter of cylinder (inside diameter of 
liner) inches; f = 4000 (f is large as liners are made of special iron, 
and usually have steam pressure outside) ; C = constant, to allow for 
reboring=.5 for large cylinders, =.25 for less than 20" diameter. 

For barrels: 

When liner is used, D = d + 2t + 2c. (166) 

d = diameter of liner (inside) inches. 

t = thickness of liner. 

c = width of steam space around liner = about f". 

PD 

T = thickness of barrel= -^ . (167) 

Prz boiler pressure. 

D = diameter of barrel inches. 

f-L = 2250 for vertical cylinders. 

The thickness of the remaining parts of the cylinders are calcu- 
lated by taking certain percentages of a constant K as follows : 

K = -|5- + .01D. (168) 



I. H. P. KEQUIRED FOR GIVEN VESSEL. 223 

Thickness of various parts of cylinders using above constant : 



Part. 

steam Ports 

Cylinder bottoms, double 
Cylinder bottoms, single 

Cylinder flanges 

Cylinder covers, double . 
Valve chests 



H. P. 



I. P. 



L. P. 



K, 



X .65 
X .7 
X .85 
xl.2 
X .65 
X .65 



K/ 



X .65 
X .7 
X .85 
xl.2 
X .65 
x .65 



Kx. 



X .66 

X .7 

X .85 

X 1.2 

X .66 

X .65 



The table above represents the practice of the Bureau of Steam 
Engineering for a general guide, but it gives rather heavy results, 
particularly for the thickness of valve chests. 

For valve chests use K X .58 ; this will give more reasonable 
thickness. 

In formula (168) D is found by formula (166). The following 
table gives values of the pressure P used in formulae (165), (166), 
(167) and (168), in the latest Bureau design. 

Steam pressures to use in preparing design. At H. P. valve chest 
265 pounds p. g.= (280 pounds abs.). 100° superheat: 



Pressures. 



H.P. 


I. P. 


L. P. 


425 


200 


100 


275 


130 


40 


265 


126 


100 


265 


100 


35 



Con- 
denser. 



Test pressure 

Cylinder relief valves set at. . . 
For cylinder liners and barrels 
For load on piston 



80 



When the covers are steel use a value of f^ of 9000. 

The thickness of small cylinders, having no liners, is calculated 
in the same way, and a small allowance added for wear, reboring, 
etc. 

From practical considerations of stiffness, and to insure reliable 
castings, the I. P. details are usually made about the same thickness 
as the H. P. The L. P. details are somewhat less, in the case of 
large modern engines about J inch less. If the solution of the 
formula gives much smaller values for L. P. cylinder than about 
■J inch less than I. P., then the thickness should be arbitrarily 
increased. 



2,24 ENGINEERING MECHANICS. 

Very frequently the inner wall, that is, the wall on which the 
cylinder pressure acts, of double bottoms and covers, is made slightly 
thicker than the outer wall. When this is done it is usual to make 
the outer wall -J inch less than found by the formula, and to add 
i inch to the inner wall. 

When double covers and bottoms are used they are strengthened 
and stiffened by being connected by radial webs. The height of 
radial webs in single covers, or bottoms, and the height of the clear 
space between walls in double covers and bottoms is given by 

hz=7t to 9t, (169) 

t being the thickness of wall. 

The thickness of webs may be made three-quarters of the thick- 
ness of cylinder liner, or thickness of web = . 75 x thickness of liner. 

Determine the number of webs by the following: 

D + 20 



Number of webs : 



H. P. 
I. P. 
L. P. 



9 

D + 20 where d = dia. of corresponding 
10 cylinder. 

D + 20 
11 



When single covers are used, the material is generally cast steel, 
giving a light cover, but when the cover is double it is necessary 
to use cast iron on account of the rather complicated form. The 
double cover also affords a convenient method of steam jacketing 
the cylinder ends. Double covers and bottoms are used in all large 
modern naval engines. 

128. Cylinder-head Bolts or Studs. — For the method of deter- 
mining these see Art. 45. 

129. The Pistons. — The H. P. piston of modern engines is made 
of cast iron, the I. P. and L. P. pistons of cast steel. 

Cast iron is used for the H. P. because it is desired that the 
weight of the reciprocating parts for each cylinder should be as 
nearly as practicable constant, in order to obtain a well-balanced 
engine. Therefore, if steel were used the piston would still be made 
much heavier than necessary from considerations of strength only, 
and there would be a waste of money in using the more expensive 
material. For a good example of a modern piston see Barton's 
Plate IX. 



226 



ENGI]^EERING MECHANICS. 



The slope of I. P. is then tan"^ ^ = about 16°, then c=:.725. 

i = 4.125x. 725 = 2.99, or 3". 
a = .45xi = .45x3 = 1.35, or about If". 

The method of design given above is based on the general formula 
for a circular plate, supported at the center and uniformly loaded. 




-1?= €/ia/n. of Cy/i/?aer- — 



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20 -^ GO SO /OO /ZO MO /60 /60 200 220 240 260 280 300 320 

Fig. 112. 



This formula in its general form is t=CxDVp. The modifica- 
tions are in the values of the constant based on the coned shape 
given the piston, considerations of inertia and practical allowances. 
Proportions of cast-steel pistons (Bauer) : * 



* Curves plotted from Bauer's tables. 



228 



E^^GL^^EERIXG MECHANICS. 



equal bearing pressure between ring and cylinder wall, and also to 
hold the piston centrally in the cylinder. 

130. Piston Rods. — The formula used by the Bureau of Steam 
Engineering is: 

fXS 



1+^a 



(172)* 



P = total load on piston rod in pounds. 
S = sectional area of rod sq. in. 
a = 1/2250. 

l-=r=usually 10 to 12. 

f = 4500 (steel) to 6500 (high-gi-ade steel), 




Fig. 114. 



The rods are made of "high grade ^^ steel (nickel steel) and are 
usually hollow to facilitate tempering. When the rods are to be 
hollow, solve for a solid rod and find the diameter of the equivalent 
hollow rod, by the formula 



d^ (solid) 



d,^. 



di 



■d/ 



the size of the hole being known or assumed. The usual practice is 
to make the diameter of the hole = -| the outside diameter of the rod. 
Applying this formula to the I. P. and L. P. rods, generally gives 
smaller sizes than for the H. P., but the I. P. and L. P. rods are 
actually made of the same dimensions as the H. P., except that the 



* This will be recognized as a modified form of Gordon's Strut 
Formula. 



I. H. P. EEQUIEED FOR GIVEN VESSEL. 229 

diameter of the hole of the L. P. rod is sometimes made larger than 
the H. P. and I. P. 

Collars. — The rods are forged with a collar^ sketch in a previous 
note^ and is flush with the inside, or lower face of the piston when 
set up. The diameter of this collar at the piston end of the rod is 
made 1.2 x diameter of rod, and the thickness of the collar is about 
.15 X diameter of the rod. 

Taper of Rods.— Where the rods pass through the hub of the 
piston they are given the conventional taper of l-J inches to 1 foot. 
The -end of the rod projecting above the piston is turned down and 
threaded for the Bureau standard slotted nut. The diameter of the 

end is such that area at bottom of thread = ^ . ^\^ . A shear- 

ll;,oUO 

ing stress, tending to strip the threads of the nut, of 11,500 pounds 

per square inch is allowed; this makes the height of the nut = the 

nominal diameter of the threaded part of the rod. 

The standard proportions of the nut are tabulated by the Bureau, 
but, in case the table is not available, the following rough approxi- 
mation may be used: A = 1.7d; B = .2d; C = .ld; D = d. 

131. Details of H. P. Cylinders, Valves, etc. — For purposes of 
illustration the details of an actual modern engine of about 12,000 
L H. P. is used. This engine has the following principal dimen- 
sions, viz: Cylinder diameters — H. P., 38^ inches; LP., 63-1 
inches; two L. P., 74 inches (each) by 48-inch stroke; E. P. M., 
120. 

Shape of Bottom Heads and Covers. — This is determined by the 
shape of the piston, and is so arranged that the clearance is kept 
uniform at -J inch at the top and f inch at the bottom, these clear- 
ances being called for by the specifications and based on expe- 
rience. The opening in the bottom head for the piston-rod stuffing 
box is determined by the kind of packing to be used, and is made to 
conform to the sizes given by the makers catalogue : in this case 
12 J inches for the upper (or inner wall) and 13^ inches for the 
lower (or outer wall), so that the stuffing box can be entered into 
place from below. The diameter of the flange or boss, for securing 
the stuffing box is 18 inches. Fig. 2, Plate VII, Barton^s " I^aval 
Engines and Machinery," shows a high pressure C3dinder similar to 
the one under discussion. 

Counterhore. — The cylinder liner is counterbored at each end to 
a diameter of 39 inches, which is ^ inch greater diameter than the 



230 ENGINEERING MECHANICS. 

bore of the liner, and the length of the connterbore at each end is 
snch that the piston ring overruns ^ inch at the top and -J inch at 
the bottom. The object of this connterbore is to prevent wearing 
shoulders at the ends of- the stroke. The exact position of the 
connterbore is found by drawing the piston in position with the 
proper clearance, when the length of the connterbore can be laid 
off. 

Note on Velocity of Steam. — The sizes of the cross-sectional areas 
of steam and exhaust passages, receiver pipes, etc., are determined 
by fixing a maximum rate of flow of steam, which must not be ex- 
ceeded. In ordinary engines these limiting velocities are : 

1. Main steam pipe, v = 6000 to 7500 feet per minute. If the 
steam pipe is very long, the velocity should be taken a little less. 

2. In steam passages of — 

H. P. cylinder, v=5000 to 6000 feet per minute. 
I. P. cylinder, v = 6000 to 7000 feet per minute. 
L. P. cylinder, v = 7000 to 8000 feet per minute. 

3. In exhaust passages and receiver pipes of — 

H. P. cylinder, v=4000 to 5000 feet per minute. 
I. P. cylinder, v=5000 to 5500 feet per minute. 
L. P. cylinder, v = 5500 to 6500 feet per minute. 

4. The velocities given above are increased from 10 to 20^ in 
very light high speed engines, where the saving of weight and space 
is of more importance than economy at the highest speeds. 

5. In naval engines these velocities are often somewhat exceeded. 
Port Area for H. P. Cylinder. — Since the steam has to exhaust 

through the same port by which it enters, the size of the ports must 

be governed by the velocity of the exhaust steam, rather than by the 

velocity of steam at admission. For the H. P. this area then must be 

such that the velocity of exhaust does not exceed 5000 feet per 

minute (see note on velocity of steam). Design the valve to open 

wide for exhaust. 

. . , _ Area of H. P. cylinder x piston speed 

Area ot port- ^^ 

Note. — The numerator of this faction represents the volume 
swept by the H. P. piston per minute. 

In this case area of port = |^^ (^^^)'x ^-^^-^^^^-W 5000 = 223.5. 



233 



ENGINEERING MECHANICS. 



ton, Fig. 37. The Bureau of Steam Engineering specifies that the 
bridges must take up not more than from 20^ to 25^ of the opening, 
leaving from SO fo to 75fc of clear area for the passage of the steam. 

Diameter of H. P. Valve. — This diameter must be such as to give 
an area equal to that of the ports. The maximum port opening for 
steam is assumed in this calculation, and for an engine of this size 
experience shows that about 3 inches is right. 

A Zeuner valve diagram is now constructed. G-iven length of 
connecting rod, 8 feet. Stroke, 48 inches. Travel of valve (by 
specification) lOj inches; assume, lead (angular), 10°; top piston 



Top ^e/et 







Sca/e, 

i^c7/ye C/rc/e /r//^zJ. 



Top/-e/7c/ /OZ 



/Release 



^o6i^om O^^ O// 7^Z 



Fig. 115. — Zeuner Valve Diagram. 



position at cut-off, .78 of stroke. To allow for the angularity of the 
connecting rod take the top cut-off at 78^ of the stroke, and the 
bottom cut-off at 72^, giving a mean value for both strokes of 75^. 
Eelease at ^^ of the stroke from each end. These assumptions are 
based on experience gained from previous engines, and from a gen- 
eral knowledge of practical requirements. Then the following points 
and dimensions are obtained from the diagram. 

Steam lap, top = 2yy , bottom = 2" , exhaust lap top ( — ) ^' bot- 
tom 0. Maximum opening to steam top, 2V9, bottom 3J'', exhaust 
opening, top = 5|", bottom 5 J". 

In order that the total length of the valve chest may be kept 
within practicable limits, the length of the opening in the liners will 



I. H. P. REQUIEED FOR GIVEN VESSEL. 



233 



be taken as 4 inches. The circumference of piston valve X length of 
opening X per cent of clear openings area of steam port. Or 
7rDx4x.775 = 223.5, 
223.5 



D = 



22.95, or 23''. 



7rx4x.775 

WorHng Liner. — There will be at each end of the valve chest a 
working liner made of hard close-grained cast iron, as hard as can 
be worked. These are forced into place and held by screws tapped 



— je: 




Center L/ae o/ 

Fig. 116.— Top Valve-Chest Liner and Steam Port. 

half into the liners at the ends and half into the metal of the cast- 
ings. See Barton, p. 86. The upper liner is driven into place from 
the top, the lower liner from the bottom. The outside diameters of 
these liners are turned down in steps, as shown in the sketch so as to 
enter and seat accurately (see Fig. 116). 

The thickness of the liners is found from the formula i— "oTT^" 
where D = internal diameter of liner and t = thickness, both in 
inches. Thus 



+ =23 
^ "TO" 



.5 = l'/27 say I3V. 



234 



ENGINEERING MECHANICS. 



The thickness of the valve-chest casting has been determined by 
the formula given previously. 

The narrowest part of the pocket surrounding the liner and 
which forms part of the steam post is 1-| inches in the direction 
of the diameter of the chest, and 7 inches vertically. 

Fig. 116 shows the method of fitting the liner in the casting, 




Fig. 117.— Section on ABCD (Fig. 116). 



together with the dimensions used. It will be noticed that ample 
clearance for entering the liner through the end of the casting is 
provided, the diameter of the opening being 26J inches. Note that 
in the sketch all webs have been omitted for the sake of clearness. 
The lengthwise dimension of the port, where it runs vertically, has 
been reduced to 6 inches. This may be done without materially 
decreasing the area, as the width at this point has a circular form, 
surrounding the cylinder wall. 



I. H. r. REQUIRED FOR GIVEN VESSEL. 235 

The distance between the center lines of the H. P. cylinder and 
the H. P. valve chest depends upon the length of the crank pin and 
the main bearings^ and the thickness of the crank web. These have 
been already calculated, and the items are added as follows: | 
length of crank pin + width of crank web + clearance + length of 
main bearing + distance from edge of main bearing to vertical center 
line of go-ahead eccentric. This works out, for the engine used as 
an illustration, as 4' — f". 

It is desirable to keep this distance as small as possible, in order 
to reduce the clearance volume, which is always a large percentage 
of the piston displacement, especially in the H. P. cylinder. 

Port in Valve Chest. — To determine the shape and size of the 
port immediately surrounding the valve-chest liner, proceed as 
follows : The dimensions of the port at the cylinder walls, as pre- 
viously found, are 6^ inches by 35 inches, giving an area of 223.5 
square inches. Eef erring to the sketches, Figs. 116 and 117, it is 
seen that each quadrant of the liner must furnish one-fourth of 
the steam; hence the sectional area at xx (each side) must be equal 
to one-fourth of 223.5 square inches, or 55.9 square inches. The 
depth of the port around the liner has been made 7 inches, x then 

= -^ =8 inches (nearly). The outline of the passage can now 

be put in, using arcs of circles and straight lines and rounding o3 
the corners. 

Note. — The value of x of 8 inches as found above, was consider- 
ably reduced in this case, in order to keep down the clearance 
volume, even at the expense of a local increase in the velocity of the 
steam, between the bridges of the liner. This is a good example of 
one of the many compromises necessary in designing machinery. . 

Length of Working Faces of Valve and Liner. — The length of the 
valve must equal the opening of the port in the liner + steam lap 
-f- exhaust lap. In this case, from the Zeuner diagram previously 
constructed, length top = 4:-f 2fe--f ( — J) =5yV'> length of bottom 
disk = 4 + 2 + = 6. The length thus found is that of the valve rings. 
For the length of the liner we now know the travel of the valve and 
the length of the valve face, so the length must equal the sum of 
these, or 10^ + 6 = 16^. Each end of the liner is counterbored, or 
chamfered off ^ inch to allow the valve rings to overrun in order to 
prevent the formation of shoulders (see Fig. 116). 



236 ENGINEERING MECHANICS. 

To Fix the Position of the Liners in the Valve-chest Casting. — 
The total length of the valve chest is made the same as the total 
length of the cylinder. See Barton^ Plate VII, Fig. 2. In order to 
reduce the length of the steam passages to each end of the cylinder 
it would be well to separate the two liners, axially, as far as possible, 
but here again a compromise must be made in order to allow space 
for the opening of the exhaust nozzle to the I. P. cylinder, the valve 
taking steam at the middle and exhausting at the ends.* The liners 
are placed equally distant from the transverse center line of the 
cylinder and valve chest. The lower valve-chest cover carries the 
valve-stem stuffing box and the upper cover carries the balance 
piston cylinder, and these project into the valve chest in the manner 
shown by Fig. 37, p. 87, Barton. 

The size of the exhaust nozzle will be found to be about 14 inches 
internal diameter, and this, together with the necessity of allowing 
sufficient clearance between the valve and the lower edge of the 
balance cylinder, fixes the highest point at which the upper edge of 
the upper liner can be located. This is 15 J inches from the upper 
flange of valve chest. The total length of the valve chest is 6 feet 3 
inches, or 37^ inches measured each way from the middle line. So we 
now have 37^— (15| + 16J) =5| inches. This 5f inches locates the 
inner edge of the liner, from the middle lines. As shown in Fig. 37, 
Barton, the edges of the openings in the liner are not flush with the 
edges of the ports in the casting, in order to allow of slight altera- 
tions should any be found desirable after the engine has been tried. 

Valve. — The valve can now be laid off in its middle position, the 
principal dimensions being taken from the diagrams previously con- 
structed. The steam enters at the middle, and as there are to be 
two exhaust nozzles, one on each side, at the upper part of the valve 
chest, the exhaust from the bottom of the cylinder must pass 
through the valve itself. Therefore, the valve will be similar to that 
shown by Fig. 37, Barton. 

Valve Stem. — The specifications require as follows : " The high 
and intermediate pressure valve stems will be 3J inches in diameter 
at the stuffing boxes, reduced to 2J inches where they pass through 
the valves and to If inches above the valves. The balance pistons 

* In the latest engines the length of the valve chest has been con- 
siderably increased in order to locate the valve liners nearly on a line 
with the top and bottom of the cylinder, thus giving an almost straight 
passage and considerably reducing the clearance volume. 



T. H. P. REQUIRED EOR GIVEN VESSEL. 237 

will be secured to the valve stems by l|-ineli nut?, as sliowji/' (See 
Barton, Fig. 37.) 

Valve-stem Stuffing Box. — ^The opening in the lower valve-chest 
cover will be made to accommodate a reliable make of stuffing box 
for a 3J-inch rod. United States metallic packing is very generally 
used. 

Cylinder Cover and Valve-Chest Cover Studs and Flanges. — Expe- 
rience shows that the diameter of the cylinder cover studs for the 
pressure used in this engine should be somewhere about 1^ inches. 
7'o find the pitch circle and the number of studs. The thickness of 
the cylinder liner has already been found, which doubled and added 
to the inside diameter with a further addition for the seatings of 
liner, gives, in this case, 42| inches. The clear entrance at the top 
of the cylinder must be made somewhat larger than this maximum 
diameter, to provide clearance for entering the liner. Make this 
43J inches. The width of the flange should be from two and one- 
lialf to three times the diameter of the stud, or U" X U" - 3f ". This 
makes the outside diameter of the flange 43J-f 7^^ = 51 inches and 
the diameter of the pitch circle (the mean between inside and out- 
side diameters) (51 + 43J) -^2 = 47J. The pitch is roughly taken 
at between 4 and 5 inches in order to make a steam-tight joint. For 
steel studs of 1^ inches diameter a working stress of 7500 may be 
allowed. Then for a first approximation. Pitch = 4 and number 

of studs - — -r^^^^ =37.1, or sav 36 bolts. To determine the exact 
4 

diameter of studs: 

-X (41.25)^ X365 = 36X ^f X 7500, 
4 4 

cl^= ^^-~^^|^ d = 1.48 (effective) or use 36 ir^tuds. 
36x7500 

And for convenience the same size will be used for all C3dinders. In 
the same way the studs for valve-chest covers are found. A stud 
not less than 1 inch should be used, so the width of the flange will 
be from 2J to 3 inches. Using the larger value, and the diameter of 
valve chest previously found, viz, 26|^ inches, the diameter of the 
pitch circle will be 29^ inches. 

-^J^i^^ :=23.2 or use 24 studs. 
4 

16 



238 EXGIXEERIXG MECIIAXICS. 

Note. — The allowable stress per square inch decreases with the 
decrease in size of the bolt, and in this case, since the valve takes 
steam on the inside, the pressure on the cover for calculation is 
125 pounds per square inch. 

Then 

4- X(29.5)-Xl25 = 24x ^ X 5000, 
4 V ^ 4 

d'= -4^f^^^^^ , d=r.952 (effective) or use 24, 1^" studs. 
24X£>000 . ^ ^ 

Valve-Chest Cover. — A cover of this kind is considered as secured 
around the edge in a manner between merely " supported " and 

"fixed," so that the formula may be of the form t= J ^^P , 

t= thickness in inches, r=: radius of cover in inches (take radius of 
pitch circle of bolts or studs) ; p = maximum pressure in pounds per 
square inch; and f the greatest permissible stress of the material. 
For cast-iron covers, when a suitable value of f is inserted and a 
constant added for practical. purposes, this becomes 

t^y/ gx^^xp T+.5. 

V I 4x45,000 J 
This will give satisfactory results for dished covers, with ribs. The 
thickness of the ribs is made .9xt, and the number found as given 
for cylinder covers. 

This same formula may be used for the cover of the manhole in 
the cylinder cover. 

Intermediate Pressure Cylinder. — Diameter, 63|- inches; area 
= 3170 square inches. The thickness of liner, of barrel, of bottom 
head and cover are to be calculated by the rules and formula already 
given. Also the piston, piston rod, etc. 

Valves. — In order to save length and to keep the diameter of the 
valve within reasonable limits two piston valves will be used, the 
stems being connected to a common yoke. (See Barton, Plates I, II 
and III.) 

Dimensions of Steam Ports at I. P. Cylinder. — For this cylinder 
the width of the port may be taken slightly larger than the bore, 
which is 63 J inches, say make the width 66 inches. 

. _ Area I. P. cylinder x piston speed 
^''^- 5500 

5500 = 552. 



_ /ttX (63^ ^120x48x2 \ 
" I ~^^~ ^ I2 i • 



I. PI. p. REQUIRED FOR GIVEN VESSEL. 239 

552 

Then length of port= — - =8.36 inches. 'Now since the cor- 
responding dimension of the H. P. has been fonnd to be 6J inches, 
and since the facings for the stuffing boxes mnst be in line for all 
cylinders, and the npper flanges of cylinders must also be in one 
horizontal plane, it will be impracticable to use so long a port as 
found above. In the engine under consideration this waB reduced 
to 7 inches. Using this valve the port area becomes 66x7 = 462 

3170x960 
square inches, and the velocity will then be ' —^ — = 6580 feet 

per minute. As this falls within the limits for steam velocities in 
I. P. cylinder, it may be used without fear of unduly increasing the 
back pressure on I. P. piston during exhaust. 

Diameter of I. P. Valves. — Make the length of port in the valve 
liner 4 inches, as for the H. P. and assimie 20;^ of the opening to 
be taken up by the bridges, leaving 80;/ clear opening. Then, since 
the port area has been decided as 462 square inches, and there are 
to be two valves, 2(7rdx4x.8) =462, or d = 22.9, say 23". Then 
the thickness of liners, and the inside diameter of valve chest, and 
outside diameter of valve-chest flange will be the same as for the 
H. P. 

Cylinder Cover Bolts. — The diameter of these, for the H. P. was 
decided as IJ inches and for convenience the same size will be used 
for this cylinder. Calculating the thickness of the I. P. cylinder 
liner by the formula previously given, it is found to be 1-J inches, 
therefore, proceeding as in the case of the H. P., the width of flange 
should be the same, or 3f inches, and the inside diameter of 
cylinder, at flange 6TJ inches and diameter of pitch circle = 71. 
inches. Then. 

4 4 

Number of bolts = n = 37.8, sav 38; pitch = ''^^-^ =5.87". This is 

38 

too wide a spacing to make a reliable joint, and although 38 If-inch 

bolts are enough for strength, the number must be increased to make 

a tight joint. Assume a pitch of 4 inches. Then 

n= ""^^^ =55.8, or 56, use 56 bolts. 
4 

From the calculation for the H. P. the diameter of valve-chest 
cover (outside diameter of flange), is 32^ inches. 



2^0 



EXGIXEERING MECHAjSTICS. 



To Fix the Position of the Axes of Valve Chests relative to Axis 
of I. 'P. Cylinder. — La}^ off the cylinder representing the outside 
diameter of the I. P. cylinder cover, the diameter in this case being 
74J inches. Now the distance between the centers of the two valves, 
in the athwartships direction is often taken at from 1^ to If of the 
diameter of the valve itself. This gives good proportions for the 
passages connecting the two valve chests, bnt the exact dimension 
is not of great importance, so long as it is sufficient to give space for 
the rather complicated core work in moulding the casting, and at 
the same time is not so great as to unduly increase the clearance 
volume. Also it is desirable to keep the length of the yoke connect- 
ing the lower ends of the valve stems as short as practicable. Taking 




Fig. 118. 



IJ times the diameter of the valve gives IJx 23 = 40.25 inches, or 
say 40Jinches. 

Therefore, the center of each valve is 20^ inches from the longi- 
tudinal center line of the engine. The minimum distance longi- 
tudinally from the center of the cylinder to the transverse line con- 
necting the centers of the two valves, is fixed, as in the case of the 
H. P. by the length of shaft necessary for crank pin, width of 
crank web, length of main bearing, etc., up to the center of I. P. go- 
ahead eccentric. Lay off' this minimum distance AB on the fore 
and aft center line of the engine, and draw a transverse line 
through the point B. On this transverse line, from B, in each 
direction lay oft' 20J inches, giving the points C and D. With C 
and 13 as centers, and with radii equal to ^ the diameter of valve- 
cliest covers, strike circles. If these circles clear the circle of the 
cylinder cover by, say 3 or 4 inches, the distance AB thus found may 



1. H. P. REQUIRED FOR GIVEN AESSEL. 241 

be used. The clearance of 3 or 4 inches is allowed for convenience 
in getting a wrench on the nuts of the cover studs. If the circles 
of the valve covers cut the circle of the cylinder cover, then the 
point B must be moved back from A, until the valve-chest covers 
clear the C3dinder cover by 3 or 4 inches. 

Proceeding as above in the case used; the diameter of the LP. 
cylinder cover is 74f inches. The minimum possible length for AB 
is 4 feet Of inches. The diameter of the valve-chest covers is 32^ 
inches. Using these dimensions it is found that the valve-cover 
circles cut the cylinder-cover circle. Therefore, B must be moved 
back. By actual trial it is found that AB must be given a value of 
4 feet 5i inches, in order to give a clearance of 3^ inches at E. F. 

Loiv-Pressurc Cylinder. — The details of this cylinder are worked 
out in a way exactly similar to that given for the H. P. and I. P. 
The diameter of each L. P. cylinder in the engine used for illustra- 
tion, is 74 inches. The diameter of the upper cylinder flange, and, 
therefore, of the cylinder cover is 85J inches. The diameter of each 
L. P. valve, there being two piston valves, for each L. P. cylinder 
works out at 28^ inches, using- a steam velocity of 7500. 

NoTE.^ — In the engine used in this illustration, the steam veloci- 
ties used were higher than those given in the note on velocity of 
steam. The engine was evidently designed by using the velocities 
allowed in steam passages. 

The transverse distance between centers of the two valve chests 
is 45 inches, and the distance, fore and aft, between centers of valve 
chests and center of cylinder is 5 feet 2 inches. 

Distance between Cylinders. — These are given by the specifica- 
tions: F. L. P. to H. P., 5 feet 10 inches; H. P. to I. P., 12 feet 
6 inches; I. P. to A. L. P., 6 feet 11 inches. 

Steam Pipe, Receiver Pipes and Exhaust Pipes and Nozzles. — 
The specifications call for the main steam pipe to be connected to a 
nozzle at the middle of the H. P. valve chest, on the inboard side of 
the engine. There are nozzles, at the upper part of the H. P. valve 
chest on both the inboard and outboard sides for the receiver pipers. 
For the I. P. there are nozzles, on the upper part of the I. P. valve 
chests, on both the inboard and outboard sides for the receiver pipes 
from the H. P. There are also, from the middle of the I. P. valve 
chest, on the outboard side, two nozzles, one for the receiver pipe 
leading to the F. L. P. and the other for the receiver pipe leading 
to the A. L. P. As these pipes approach the L. P. valve chests they 



242 ENGINEERING MECHANICS. 

divide into two branches, thus leading the steam to both the upper 
and lower parts of the L. P. valve chests. The L. P. chests will 
then have a nozzle at the upper part and another at the lower part 
to receive the steam. They, also, each have a large central nozzle, 
on the outboard side for the main exhaust pipes. All of these steam, 
receiver and exhaust pipes have slip joints. 

To Find the Sizes of the Pipes. — The velocity of steam through 
the main steam pipe may be 7500 feet per minute. 
Then 

area H. P. cylinder X piston speed « . 

^-^-^7 — ^-5-!- i = area of pipe. 

velocity of flow ^ ^ 

The cut-off in the H. P. cylinder is not considered, and the size of 
the pipe is taken as if the steam followed full stroke. Then 

— — ^r — %f^p.p, ==148 square inches, or dia. = 13| (from table). 

Make main steam pipe 14 inches diameter. 

Diameter of First Receiver Pipe, and of the Nozzles to which it 

Connects at H. P. and I. P. Valve Chests. — Considering this pipe 

as the exhaust pipe from the H. P. cylinder, it must have an area at 

least equal to the area of the exhaust port opening of the H. P. 

cylinder. This area was found to be 223-1 square inches. Since 

there are two receiver pipes, each must have an area of at least 

223 5 

— - =111.75 square inches, or the diameter must be 12 inches 

(nearly). These pipes also act as steam pipes for the I. P. cylinder. 
Treating them as such 

. 4? 1 • area I. P. cvlinder x piston speed 

Area of each pipe = r- r^^; ^ — 

steam velocity 

_ vrX (631)^X960 _ ^ ^ 

- 4 X 7000 X 2 ~ ^ ' ^^^^^^ ''^'^^'• 

And the corresponding diameter is 16f inches nearly. The size is 

to a certain extent governed by the opening to exhaust of the H. P. 

cylinder, so that two 16f-inch pipes are larger than necessary. Take 

the diameter of the pipe part way between the two solutions, about 

a mean between, the two, disregarding fractions. Therefore, make 

each pipe 14 inches diameter. 

Size of Second Receiver Pipes. — The diameter of these pipes is 

determined in a manner similar to that used above. Thus, the area 

of exhaust port in I. P. cylinder has been found to be 552 square 



I. H. P. PtEQUIRED FOR GIVEN VESSEL. 



243 



inches, and since tliere are two exhaust pipes, one leading to the 
F. L. P. and the other to the A. L. P. cylinders, the area for each 

is ^^ = 276 square inches, the corresponding diameter being 18f 

inches. Considered as the steam pipe for the L. P. C3dinders, and 
allowing a velocity of flow of 8000 a pipe about 25 inches diameter 
will be required. Taking a valve between, as before, the diameter to 
be used is 20 inches. These pipes divide, as they approach the L. P. 
cjiinders, to lead the steam to both ends of the L. P. valve chests. 
In order to maintain an equal area, each branch must be between 
14 inches and 15 inches diameter. Make each branch 15 inches 
diameter. 




Pig. 119. — Sketcli Arrangement of Steam, Receiver and Exhaust Piping. 



Main Exhaust Pipe. — The exhaust from each L. P. cylinder is 
led to separate nozzles on the condenser. The area of the L. P. 
exhaust port is 7^ X 71 inches giving an area of about 533 square 
inches, the corresponding diameter being slightly over 26 inches. 
The size of exhaust pipe actually used is 27 inches. 

Laying off Pipes and Nozzles. — Having now determined the 
necessary sizes of the steam, receiver and exhaust piping, the sizes 
and angles of the nozzles on the several valve chests to which these 
pipes are connected can be laid off. Make a plan of the complete 
engine, showing the cylinders and valve chests in their correct rela- 
tive positions, as in Fig. 119. 

The main steam pipe, A, is led along the inboard side of the 
engine, as shown. B represents the main throttle valve, and C, the 
nozzle for steam of the H. P. valve chest. DD represent the H. P. 
exhaust nozzles, and EE the I. P. steam nozzles. FF are the I. P. 



244 ENGINEERING MECHANICS. 

exhaust nozzles. GG- the receiver pipes to the L. P. cylinders, and 
HH the steam nozzles of the L. P. valve chests. II the L. P. exhaust 
nozzles. Having thus obtained the general arrangement of the 
piping, the several nozzles are designed in detail on the dravdngs 
for the cylinders. 

This practically finishes the design of the cylinders, there remain- 
ing only the flanges for bolting the various castings to each other, 
and the feet for attachment to the engine columns, or framing. In 
addition the bosses for attachment of the receiver safety valves, 
c^dinder relief valves, drain valve connections, indicator bosses, peep 
holes for observing the setting of the main valves, etc., remain. 
These will not be described here. 

Donhle-ported Slide Valve for L. P. — Quite frequently a double- 
ported flat slide valve is used for the L. P. engines instead of two 
piston valves. These are described in Barton's " Naval Engines and 
Machinery/' P- ^1? a^d '^^^ valve diagram with dimensions of the 
parts for such a valve is worked out in the plates (see Plate 
XXXVII). 

132. Check for Strength of Cast-Steel Pistons such as are Used 
for I. P. and L. P. of Naval Engines. — After the piston has been 
designed and the thickness at different points determined, either by 
a method similar to that described, or by using the design of some 
other engine approximately similar to the new engine, the strength 
is checked by the following method: 

The rupture of a piston is caused by the force P acting normally 
to the face. Eupture is resisted by the section of the piston made 
by the plane YY through the axis of the rod. To find the combined 
bending moments of the resolved forces, moments are taken from 
the center of gravity of the section lying in the plane YY. 

The resolved force parallel to the rod (in the direction M^) is 
the pressure Pxthe projected area of the piston = PX7rE^, 

On each side of the piston this force is ^^^ and it acts at the 

center of gravity of a semicircle =.4244E from YY, hence its 

moment = M, = P X ""f^ X .4244E = .21227rE^P. 

The resolved force in the direction M2 is P X projected area of the 
section of the piston = 2 (triangle ABD + rectangle BCED)xP 

=zPx2(^Xr + h,xr)-Pr(h-l-2h,). 



I. H. P. REQUIRED FOR GIVEN VESSEL. 



245 



Note. — The slight flat part under the hub is disregarded, and 
the form considered as being conical, as if the line of the inner face 
continued unbroken to A. This assumption does not cause an 
appreciable error. In the same way, as will be seen later on, the 
hub is disregarded and the line of the outside surface considered 
unbroken to the axis. 

If yc = distance of the center of gravity of the section from the 
base EC of the piston, then the arm of the resolved part of the ring 

force (projected rectangle CBDE) is Vc — ~, and the moment of 



this force=Px3hiXrx(yc- ^j=Prx 3h,(y„ 



(A) 




Fig. 120. — Piston (Calculations for Strength), 



The arm for the conical part of the piston is at the center of 

gravity of the projected triangle or at -Jh above BI) = \-\-^ above 

o 



CE, or is jc- f hi + ^-j distant from the center of gravity of the 
metal section of the piston, and the force is 2Pr — , or Prh, and the 
moment Prh [y, -(h^ + y)] . (B) 

(A) + (B)=Pr{2h,(y,-y+h[y,-(h, + |)]}=M,. 

There is also a distinct tensile pull due to the steam acting on th 
ring BC 



Px2rh 



Prh 



e 
This 



2 X area of section of piston (one side X a 

force is small and questionable and is probably accounted for in Mg. 
It is, however, added for safety. 



246 



ElS^GIis^EEETNG MECHANICS. 



The resistance to the sum of the moments is itXzx2, where 
fi = resistaiice of material per square inch, z is the modulus of sec- 
tion (one side). 

M = M,-}-M,, then M = f,x2z, f,= -^ = ^^^^\ 

To this must be added the increase for direct tension, 



Prh 



= T.-.f,= 



_ M1 + M2 



2z 



+T. 



For shearing stress take any ring at radius x and thickness t, then 
circumference of the ring = Sttx. Force on the ring = ( ttW — ttx^ ) x P 
= 27rxtfs = strength of the ring. 

. . _ P7r(B--x^) _ P( R^-x^) 
• '" 27rtx 2tx 




Fig. 121. — Showing Half Section of Piston Divided into Triangles and 
Rectangles for Finding Moment of Inertia, etc. 

Fig. 122. — Showing Triangles and Rectangles Forming Outside Sur- 
face of Piston. 



In order to find the moment of inertia and thence z, the half- 
section of the piston is divided into a number of simple figures 
(triangles and rectangles), and the moments of inertia of each 
figure about the base line are calculated; then by sums and differ- 
ences the moment of inertia of the section of the piston is obtained. 

Let k^ = square of radius of gyration of the figures 
^lyi-g-h^ for triangles, y 2-1^' for rectangles. 
b = base of figures in (2). 
h== height of figures in (2). 
a = area oi figures in (2). 
I = a X k^ = moment of inertia of figures. 
y = height of center of gravity of figures above EL. 



I. H. P. REQUIRED FOR GIVEN VESSEL. 



247 



Now, for convenience and clearness, arrange the computations in 
tabular form as follows : 





FOR OUTSIDE FORM OF PISTON 


• 




(3) 


Section. 


Shape. 


b 


h 


a 


P- 


aXk2=I 


y 


ay 


ay2 


I+ay- 


1 


Triangle. 


bi 


bi 


bihi 

2 


T^eM 


bih? 
^36~ 


yi 


aiyi 


aiy? 


Ii+aiyf 


2 


Rectangle. 


bs 


hs 


boho 


T^hi 


bohi 

12 


ys 


a2y2 


aayl 


la+asyi 


3 


Triangle. 


bg 


hs 


hshs 


^hi 


36 


ys 


azYs 


asyi 


Ts+aay? 


4 


Rectangle. 


b4 


ll4 


b4h4 


xVhS 


b4h| 
12 


y4 


a4y4 


a4yl 


l4+a4y| 


6 


Rectangle. 


bs 


b5 


bohs 


T\bi 


bshg 
12 


ys 


asys 


asyi 


Is+ajyi 


Total. 








2a 


.... 






Say 




2(I+ay2) 



FOR SPACE INCLUDED BY PISTON BEING TRIANGLE APO AND 
RECTANGLE OPQE. (4) 



6 

7 


Triangle. 
Rectangle. 


be 
b7 


be 

by 


behe 
2 

brbT 


*b? 


bebl 
36 

brb? 
12 


ye 


aeye 
ajy: 


aoyi 

ayy? 


le+aeyl 

iT+ayy? 


Total. 








2a' 








2a'y' 




2(T'+a'y'2) 



To find the center of gravity : 

-^=y. (from (3)). (a) 

:Saxy/ = Say/. (b) 

Ii = :S(I + ay2)-2ayc2 and z = i. (c) 

Jc 

Mi = .21227rE^ X P where P= initial pressure on piston. (d) 

C. of G. of internal space=^^-^ = y/ (from (4)). (e) 

Total area of cross section of (4) =2x2a'. (f) 

Load on section (4)=2x2a'xP. (g) 

Dist. (arm) between C. G.'s of (4) and (3)=yc-yc'. (h) 

M,=.(2xSa'xP)(yo-yc'). (i) 



248 EA^GINEERI?^G MECIJAXICS. 

2a'(from(3))' 
due to bending action; 






^^- 2tx ' 
shearing force at an}^ section, nsually taken at points marked + and 
c^in (1). 

To allow for shock, due to water in the cylinder, the value of f 
obtained above must be less than 8000. If it works out greater than 
8000 the thickness must be increased and another trial design made. 

In actual practice, in determining the thickness of pistons, the 
design for an engine using about the same pressures, and of about 
the same size as the new engine and which has proved satisfactory 
in service will be at hand. The pistons are then laid down with 
small changes, according to the judgment of the designer. The 
method above given is then applied as a check, and the necessary 
corrections are made if required. 

Questions and Problems. 

Describe the method of determining the I. H. P. for a new vessel 
by Kirk^s " Analj^sis." Illustrate by assuming the necessary data. 

Explain the method of determining the I. H. P. required for a 
new vessel by Froude's " Law of Comparison.^^ Illustrate by assum- 
ing data. 

]\Iake a sketch of the H. P. cylinder and valve chest of an engine 
of the following size: Diameter H. P. piston = 3 5''; stroke = 48"; 
E. P. M. = 140 ; velocity of exhaust from H. P. not greater than 
5000 feet per minute; width of piston face (total) =5J"; width of 
piston rings = 2 J"; piston clearance, top = 4'', bottom = f"; piston 
ring overruns counterbore ^", at top^ -J" at bottom; travel of valve 
= 9"; steam lap, top = 2", bottom = 1 J"; exhaust lap, top = 2J"; bot- 
tom =0"; valve to take steam on inside. Show thickness of liners, 
both cylinder and valve chest. Least distance possible between 
cylinder and valve chest axes = 3' 6". 

Design the L. P. piston (cast steel) for an engine having c3dinders 
of 381", Q^" and 2 L. P. of 74" each. Stroke = 48"; boiler pressure 
= 265 p. g. Make a sketch showing all details with dimensions and 
material used. 



I. H. P. REQUIRED EOR GIVEN VESSEL. 249 

Design the I. P. piston (cast steel) with a sketch showing all 
dimensions and material used. Data same as preceeding question. 

Sketch the upper valve-chest cover for an engine, given the fol- 
lowing data: Inside diameter of valve-chest flange = 26^"; pressure 
on cover = 125 pounds p. g. ; travel of valve=10"; diameter of 
balance piston = 6"; width of balance of piston = 2 J". Find, the 
thickness of metal (cast iron) ; number and thickness of ribs; and 
number and size of studs. 

Make a neat sketch of a cylinder cover (double) for the H. P. 
cylinder of an engine, including the number, size and pitch of the 
studs. Diameter of H. P. piston = 35"; steam pressure = 250 pounds 

Find the sizes of the main steam, receiver and main exhaust pipes 
for a four-cylinder, triple-expansion engine, from the following 
data : Distance between centers of cylinders : F. L. P. to H. P 
= 5'4i"; H. P. to I. P. = 9'10"; LP. to A. L. P. = 6' 2'' ; H. P. 
valve abaft H. P. cylinder = 3' 7''; LP. valves forward of LP. 
cylinder = 3' 7"; H. P. valve center line to transverse center line of 
I. P. valves = 2' 8"; L. P. valves forward and abaft their respective 
cylinders = 4' 9"; cylinder diameters = 32", 52" and tAvo of 72"; 
stroke = 48"; E. P.M. = 125. Velocity of steam allowed: Main 
steam pipe = 8000 feet per minute; first receiver = 7000 feet per 
minute; second receiver = 8000 feet per minute; main exhaust ]npe 
= 7000 feet per minute. Make a neat sketch showing connections 
and lead of pipes (no books) . Eeduce clearances as much as possible. 

Given: the diameter of the LP. cylinder of an engine = 58": 
E. P. M. = 125; stroke = 48". Two piston valves. Make a sketch of 
the cylinder showing steam ports, valve liners, etc. Show dimen- 
sions of parts. 

Find the I. H. P. required for a ship of 20,000 tons displacement 
and 21 knots speed, from the following data: 

Dimensions of Model Ship. Dimensions of New Ship. 

L 450 ft. 510 ft. 

B 76 ft. 10 in. 85 ft. 

Draft 24 ft. 6 in. 26 ft. 11 in. 

Immersed midship section 

area 1808 sq. ft. 2245 sq. ft. 

Speed 18 knots 21 knots 

Disp 16,000 tons 20,000 tons 

I. H. P 15,000 ? 

Find tbe power (1) by Froude^s ^^ Comparison,'' (2) by Kirk's 
" Analysis," using 5J I. H. P. per 100 square feet of wetted surface. 



250 



ENGINEERING MECHANICS. 



Find the diameters of the C3dinders of an engine from the follow- 
ing data: Triple-expansion, vertical^ inverted cylinders, total 
I. H. P. (both engines) =26,500; two L. P. cylinders each engine; 
cut-off in H. P. cylinder at 0.8 of the stroke; total ratio of expan- 
sion=:10; stroke = 48"; E. P. M. = 130; boiler pressiire = 265 p. g.; 
vacuum = 26"'. 

Find the thickness of the cylinder barrels and liners and of the 
cylinder covers and bottoms for the following engine, using Bureau 
formulae. Data : Cylinder diameters = 3 SJ", 63i" 2 of 74", by 48-inch 
stroke. Boiler pressure = 265 pounds p. g. ; first receiver = 140 
pounds (absolute) ; second receiver = 55 pounds (absolute). 

From the following data find the size of the piston rods for an 
engine: cylinder diameters = 38^", 57", 2 of 76"; steam pressure 
at PI. P. cylinder = 265 pounds p. g. ; in first receiver =110 pounds 
(absolute) ; in second receiver 45 pounds (absolute) ; vacuum = 26". 
All rods to be interchangeable. Make a sketch of the rod, showing 
threaded part for nut, collar, etc. 

Practical Problem. 
pot\'ering, size and details of cylinders, etc. 
Design, the cylinders; valve chests; pistons and rods; steam, re- 
ceiver and exhaust pipes, for a vessel of the following principal 
features : 

Displacement 20,000 tons 



Speed 

Length , 

Beam 

Draft 

Area immersed, midship section. 

Steam pressure 

Total ratio of expansion 

R. P. M 

Vacuum 

Stroke 

Cut-off in H. P. cylinder 

The following data is available: 



21 


knots 


510 


ft. 




85 


ft. 




27 


ft. 




2245 


sq. 


ft. 


265 


lbs 


per gauge. 


10 






130 






26 


in. 




48 


in. 




tV- 







Name. 


Displace- 
ment. 


L 


L 
D" 


Block 
coefficient. 


Speed. 


I.H.P. 


Ohio (class) 


12,500 
16,000 
16,000 


5.3T 
5.88 
5.6 
G.9 


16.51 
18.35 
18.35 
20.0 


.665 
.659 
.636 
.550 


18.23 
18 83 
18.79 
23.27 


17,000 


ririnnppt.ipn t, " 


20,443 




16,020 


Washingrton " 


14,500 


26,800 



I. H. P. EEQUIRED FOR GIVEN VESSEL. 251 

Use "Michigan'' as a model for Fronde's; check the cylinder 
dimensions by a ship whose engines are 32 inches, 52 inches and 
two of 61 inches, by 48-inch stroke; E. P. M., 120; I. H. P., 16,020; 
and having the same initial pressure and same total ratio of expan- 
sion as the new ship. 

Comparison. — All calculations are to be entered on the blank 
interleaved pages. 

Find (1) I. H. P. required (by two methods). 

(2) Sizes of cylinders (nse card efficiency of 53^). 

(3) Details of cylinders, liners, barrels, etc. 
(-1) Piston and piston rod details. 

Make a pencil draAving, showing H. P. cylinder and valve chest 
in section ; a drawing of the L. P. piston and its rod, half in eleva- 
tion, half in section. Such other details as time permits will be 
required by the instructor. 

Directions and suggestions for the solution of the above problem : 

In addition to the data given in the problem the following is 
needed : 

For H. P. steam ports, make widths 0.90 of diameter of H. P. 
cylinder. 

To lay down the valve liners, assume travel of valve = 10"; steam 
lap, top = 2i", bottom = 2"; exhaust lap, top= (-)J", bottom =:0"0. 
Piston overrides counterbore, at top = J'', at bottom = -J"; opening 
of port in valve liners = 4" ; inside diameter of piston rod = 4" ; use 
high-grade steel. 

Details as to length, etc., must be determined by laying down the 
work on the drawing board. 

Make steam passages as short and straight as possible. To accom- 
plish this the valve chests may be made longer over-all than the 
cylinder itself by 9 inches or 10 inches at each end. 

To Find the Diameter of the L. P. Cylinder hy Comparison.- — Let 
I. H. P., pe, S and d represent the I. H. P., mean effective pressure, 
piston speed and diameter of L. P. cylinder of a given engine, and 
I. H. P.', Pc', S' and d' the same for the required engine. Then 

LIL^ _ j)^SXd^ LH^F pe S ^^, 

I.H.P' ~pe'XS'X(dO' ^^ ~L H.P ^ pe' ^ S' 

If the initial pressure and the total ratio of expansion are the 

|-) T XT p / Q 

same in the two engines, then -^^, = 1 and (d') - = ^^~ X ^ X (P. 

Pe !• H- P- o 



252 



ENGI^TEERING MECHANICS. 



In case there are tivo L. P. cylinders, d and d' are the diameters 
corresponding to the combined areas of both L. P. cylinders. 
Solve and record results as follows : 



(1 

(2 
(3 
(4 
(5 
(6 
(^ 
(8 
(9 
(10 

(11 
(12 

(13 

(14 
(15 
(16 
(17 
(18 
(19 
(20 
(21 
(22 
(23 
(24 
(25 
(26 
(27 
(28 
(29 



I. H. P. by Fronde's law. 

I. H. P. by Kirk's analysis. 

I. H. P., mean of (1) and (2) to be nsed. 

Diameter L. P. by pe- 

Diameter L. P. by comparison. 

Diameter L. P., mean of (4) and (5) to be nsed. 

Diameter H. P. cylinder. 

Diameter I. P. cylinder. 

Thickness H. P. cylinder liner. 

Thickness H. P. cylinder barrel. 

Thickness H. P. cylinder bottom inner wall. 

Thickness H. P. cylinder bottom outer wall. 

Nnmber of ribs. 

Distance between walls, nse abont 10 inches. 

Thickness H. P. steam port. 

Thickness H. P. valve chest. 

Thickness H. P. cylinder flange. 

Area of steam ports. 

Width of steam ports. 

Length of steam ports. 

Diameter of H. P. valve. 

Thickness of valve liners. 

Piston rod, diameter solid. 

Piston rod, diameter hollow (4-inch hole). 

L. P. piston, rise of cone L. 

i and a, L. P. piston. 

h and d^, L. P. piston. 

T, L. P. piston. 

Diameter threaded part of rod. 



CHAPTEE XYll. 

Notes on the Design of Cylindrical Boilers.* 

133. General considerations in connection with the design of tlie 
ship determine the type of hoilers to be used. The type depends 
upon the vertical distance between the inner bottoms, or reverse 
frames, and the beams of the deck above; the total length and 
breadth allowed for boiler space; but principally upon the vertical 
space. 

The decision depends, somewhat, upon a tentative process. If 
the ship has small draught and is long, the small, low, or gunl)oat 
boilers may be used. 

If the ship has great beam compared with the draught, the boilers 
may be single ended, with a single fore and aft fire room. If the 
boiler space is sufficient in all respects, the power is divided among 
three or four furnace, double and single-ended, cylindrical, return- 
fire, tubular, boilers; or two furnace, cylindrical, single-ended, 
return-fire, tubular, boilers. 

The conditions limiting the diameter are as follows: Sufficient 
distance must be left between the bottom of the boiler and the 
inner bottom, or frames, to allow cleaning, painting and caulking. 
This is generally six inches for small boilers of eleven feet, or less, 
in diameter, as the rapid curvature allows these points to be reached. 
For large boilers this distance is increased to ten and sometimes 
twelve inches. The distance between sides of boilers and bulk- 
heads is about the same as the distance between shell and inner 
bottom. This is partly taken up by lagging. The distance between 
the tops of the boilers and the beams immediately above depends 
upon the spandrel space required for the uptakes. It should, when 
possible, be not less than 10 inches for boilers up to 11 feet in diam- 
eter, and as much as 15 inches for boilers of larger diameters. The 
uptakes may extend directly up through the protective deck, in 
which case no extra space is required for them, or they may join 
below the deck, as in the U. S. S. " Newark." 

Tlie distance between adjacent boilers should be the same as, or 
slightly greater than, that between the boilers and bulkheads. 

* From " Notes on Machine Design." 
17 



254 ENGINEERING MECHANICS. 

The width of fire room should be at least one foot greater than 
the length of the grates to allow the easy handling of fire tools. 

Selection of the type of holler, whether high or low, as stated 
at the beginning of this article, is determined from the class of ship 
and drauglit of water, the preference, if there is any choice allowed, 
being given to the high or return tube type, as less volume of boiler 
per I. H. P. is required with this type than with the low type. 

If the cross section of the ship will allow of boilers as large as 
11 feet 6 inches in diameter, the high boiler may be adopted; if a 
diameter of less than 11 feet 6 inches is all that can be used, the 
low type must be employed. 

This selection of type assumes that the length in the ship for 
the boilers is not determined, and the tj^pe may have to be changed 
accordingly. If the length of space for one boiler be less than 
;32 feet and more than 28 feet, the high type must be changed to 
lo^v, as the double-ended boiler is about 20 feet long and the fire 
rooms at each end must be at least 7 feet long. Single-ended high 
boilers, placed back to back, take up more length in a ship than one 
double-ended, high boiler of the same power, as space must be 
left between the backs, and also, because one single-ended boiler 
is longer than one-half of a double-ended boiler, owing to the 
full, width between the back of the combustion chamber and shell, 
and the thickness of the back heads of the boiler in each case. 

The low type must be used for lengths between 32 feet and 28 feet, 
the low boiler being at least 17 feet long and the fire room 7 feet, 
while the space behind the boiler proper must be at least 4 feet, to 
allow for uptakes. 

For boiler compartments less than 28 feet long, the single-ended 
high boiler must be used and this type requires a compartment 
length of at least 19 feet as the boiler will be nearly 11 feet long, 
fire room 6 feet and the space behind the boiler, 2 feet. 

Grate area is the first definite quantity found in designing boilers 
for a ship. It was customary at one time in our service to place in 
the ship an auxiliary boiler for running the auxiliary machinery 
while in port. These auxiliary boilers, to use only natural draught, 
were designed primarily to furnish power for a definite number of 
auxiliary engines running all the time. Auxiliary machinery is 
extremely wasteful of steam and runs under conditions of very much 
less efficiency tlian the main engines. Some auxiliary engines 
require from 30 to 40 pounds of steam per I. H. P. per hour, wliile 



NOTES OX TILE DESIGN OF CYLINDUICAI. BOIEKKS. 255 

others require from 160 to 175 pounds. If the amount of steam 
required by the auxiliary- machinery were a constant quantity, it 
would be easy enough to properly proportion the auxiliary boiler 
for its work ; but, as this cannot be done, the auxiliary' boiler has to 
suffer for the wastefulness of the auxiliary machinery and is now 
seldom or never fitted in our ships. It is customary, at present, to 
estimate the grate area necessary to furnish power for all machinery 
in operation on a trial trip, and to use this grate area as that of the 
main boilers. Where the calculations for a certain ship allow all 
boilers to be double-ended, one of these is replaced by two single- 
ended boilers, both together being of the same power as the double- 
ended boiler. These single-ended boilers are used in port for 
auxiliary purposes 

In general, the niachinerA- in operation during a trial trip is as 
follows : ]\[ain engines : air, circulating, feed, water-service and 
flushing pumps; fire-room and engine-room blowers and dynamos. 

First Method of Calculnting Grate Area. — Having determined, 
from the data obtained on different trial trips, the I. H. P. actually 
obtained per square foot of grate per hour from boilers similar to 
those proposed, tlie grate surface required is determined by dividing 
tlie given I. F. P. by the I. H. P. per square foot of grate per hour. 
The I. H. P. per square foot of grate varies, for one thing, with the 
air pressure in the fire room, and averages, for an air pressure of 
H inches of w^ater, 16.3 I. H. P. per square foot of grate per hour. 

The air pressure alloweJ by the naval authorities at present is 
for our cruisers and battleships about one inch of water. 

I. H. p. PER SQUARE FOOT OF GRATE ON TRIAL TRIPS. 
FOR AIR PRESSURE ABOUT ONE INCH. 

I. H. P. per sq. ft. 

Air Pressure. of Grate. 

Minneapolis 1.000 14.330 

Marblehead 1.080 13.160 

Massachusetts 1.000 16.900 

Oregon 1.000 18.037 

Helena 1.000 15.780 

Wilmington 1.000 15.030 

Iowa 0.987 16.010 

Maine l.OU 16.930 

xVverage for 1" '. 15.680 



25(5 ENGINEERING MECHANICS. 

FOR AIR PRESSURE ABOUT TWO INCHES. 

Texas 1.776 16.20 

San Francisco 2.020 17.92 

Baltimore 2.085 15.82 

Newark (1) 2.230 16.42 

Newark (2) 2.250 16.90 

New York 2.000 16.55 

Brooklyn 2.260 18.47 

Average 2.090 16.90 

Average of both tables above gives 16.3 I. H. P. per square foot of 
grate with an air pressure of 1^ inches. 

The data above are from official trial trips. 

Note. — Experience in the English naval service has shown the 
desirability of reducing the amount to which their boilers are 
forced, and their latest specifications for cylindrical boilers pro- 
vide a total heating surface of not less than 2.5 square feet per 
I. H. P. at natural draught power, and 12 to 12.5 I. H. P. per 
square foot of grate, while the forced draught power is limited 
to 20;^ beyond the natural draught power. 

Second Method of Calculating Grate Area. — Having determined 
from trial trips the number of pounds of coal burned per square foot 
of grate per hour, and the pounds of coal burned per I. H. P. per 
hour, the grate surface required is obtained by multiplying the 
required I. H. P. by the coal required to produce one I. H. P. per 
hour — this gives the total amount of coal burned — and dividing 
the product by the number of pounds of coal consumed per hour 
per square foot of grate. 

From 30 to 40 pounds of bituminous coal can be consumed per 
square foot of grate per hour with an air pressure of 1 inch, and 40 
to 50 with an air pressure of 2 inches. 

The amount of coal burned depends upon the kind of coal, the 
amount of pressure of draught, ratio of length of grate to diameter 
of furnace, ratio of cross-section area of tubes to grate area and 
height of smoke pipe. The height of smoke pipe has, however, 
very little effect when forced draught is used. 

Reliability of the Two Foregoing Methods. — The first method of 
determining the grate area is the more accurate. The I. H. P. 
and the grate area on trials are accurately measured, while the 
measurement of coal used is not generally made on trial trips, or if 



NOTES ON THE DESIGN OF CYLINDRICAL BOILERS. 257 

it is measured^ there is ordinarily not so much accuracy as in the 
data for the first method. 

Another method for grate area, sometimes used as a check upon 
tl;e first two methods, is as follows : 

l bs, water required per I. H. P. X I. H. P. 

Cjratearea=jj,^g^ water evaporated per lb. coal x lbs. coal burnt per 
sq. ft. grate. 

Using 21.6 pounds water per I. H. P., and 7 pounds water per 
pound of coal gives a fair approximation. From this point on the 
steps in the process of designing can be best illustrated by a problem. 

PROBLEM. 

Kequired a set of boilers to supply steam of 160 pounds pressure, 
to triple-expansion engines, such as are used in modern naval vessels 
when developing under a maximum forced draught of 2 inches of 
water, 13,500 I. H. P. for four consecutive hours. 

The following points are to be observed in the design of these 
boilers : 

(1) Stays must be placed close enough to prevent tube sheets 
springing under expansion while steam is being raised. 

(2) Stay bolts are to be screwed and the parts between the 
sheets turned down to the bottom of the thread; the screw stays 
to be screwed into the sheets and to be further held by nuts at 
both ends. 

(3) The back ends of furnaces to unite to form the lower front 
plate of the combustion chamber. 

(4) All diagonal braces to be fitted with bevelled washers where 
they intersect the plate and not to be bent. 

(5) Tubes to be pitched horizontally and vertically and not 
zio-zao^. 

(6) Corrugated or some good modern t3'pe of strengthened or 
ribbed furnaces to be used. Least inside diameter 3 feet 3 inches, 
least outside diameter 3 feet 6 inches, greatest outside diameter 
4 feet. 

(7) Diameter of tubes outside 2 J inches, ordinary tubes i^o. 12 
B. W. G., stay tubes No. 6 B. W. G. 

(8) Eatio of axial length of furnace to rectified length= 1.175. 

(9) The front head to be formed of three plates as follows : 
Portion above the tube sheet a segment of a circle, tlien bent 

back to a cvliiidrical surface. 



258 ENGINEEKIN-G MECHANICS. 

The tube sheet straight across at the top and corresponding to 
the furnace space at the bottom. The furnace sheet of the front 
head to be flanged externally to receive the furnace ends and to 
conform to the lower part of the front tube sheet. The tube sheet 
seams of these pieces to be double riveted zigzag. 

(10) Between nests of tubes there must be not less than 6 inches 
clear passage and between the sides of the combustion chamber 
and shell at least 4-J inches, this to gradually widen to 8 inches or 
9 inches at the top. 

(11) Least distance between backs of combustion chambers at 
l^ottom about 6 inches^ this to widen to about 10 inches at the top. 

(12) Manholes to be placed in each spandrel at the bottom, over 
the middle furnaces, and in upper head sheet. Principal manholes 
to be not less than 12" x 15". 

(13) The top row of tubes to be on a line that divides the cir- 
cular section into J and f, or a little above this if necessary to 
provide the proper heating surface. 

(14) A trial sketch to be made to determine the number and 
arrangements of tubes. 

(15) Two furnaces at each end to be connected to a common 
combustion chamber. 

Note. — Some of the foregoing requirements must be met by 
any marine boiler, others such as (6), (8), (9) and (15) refer to 
the particular boiler in hand. 

CALCULATION'S. 

Grate Area — Fii^st Method. — Assuming 16.3 I. H. P. per square 
foot g-rate. 



&>^ 



Grate area= ^^^- =828.2 square feet. (1) 

Grate Area — 8econd Method. — Assuming 3 pounds of coal per 
I. H. P. on forced draught trials and 50 pounds per square foot of 
grate, we have : 

Grate area = y^^55^^ =810 square feet. (2) 

Third Method. — The ^^ Yorktown's '^ water consumption as ac- 
counted for by the indicator was 17.5 pounds, per I. H. P. per hour. 
The actual water supplied was not ascertained, but supposing a 
loss from condensation, leakage, etc., of about 20^, the water used 



AZOTES ON THF. DESIGN OF CYIJNDKICAL BOILEltS. 259 

per I. H. P. = 17. 5 ^.80 1=21.8. Taking 50 pounds coal and assum- 
ing 7 pounds of water evaporated per pound coal, we have 

^ , 13,500x21.8 ^.Q . , 

Grate area = — '——- — = 848 square leet. 

As the first method is the most reliable, and in this case hap- 
pens to be about a mean between the last two, we will decide upon 
828.2 square feet as the necessary amount of grate surface. 

Size and Numher of Furnaces. — Before it is possible to fix upon 
the number of furnaces, the length of grate must be decided, for 
trial at least, and also the diameter of the furnace. Furnaces of 
cylindrical boilers should not be less than 30 inches in diameter, nor 
more than 48 inches unless in exceptional cases. Wherever possible, 
a diameter of not less than 40 inches should be given as, with smaller 
furnaces (owing to thickness of fire being practically constant 
for all diameters), the space above the fuel is much contracted and 
combustion less perfect in consequence. Long grates cannot be 
properly worked by hand and are therefore not so efficient. The 
length of grate varies from 5 feet to 7 feet and depends much upon 
the diameter of the furnace. In our service the average is between 
6 feet and 6 feet 6 inches. 

The specifications for this ship requires a furnace of 39 inches 
least inside diameter. Assuming a length of grate of 6^ feet we 

828 
have: No. furnaces == z=39-f ; taking 40 furnaces we find 

the actual length of grate = ^^^ =6.^37 = 6^ 4^ 

^ ^ 40 X 3^ 

We have, therefore, for the required boilers, 40 furnaces, 6' 4" 
X 3' 3". 

N'OTE. — The plans of the boiler space for this ship show that a 
boiler greater than 15J feet cannot be used; boilers over 15 feet in 
diameter have four furnaces if single-ended and eight if double- 
ended. We can then have ten single-ended boilers or five double- 
ended boilers, provided the boiler space will accommodate this 
number of double-ended boilers. A reference to the plans of the 
boiler space shows that four double-ended and two single-ended 
boilers can be placed. This simply amounts to calculating for 
five double-ended boilers and cutting one of them in two. In any 
case the total number of furnaces should be a multiple of twice 
the number in one head. 



260 EXGIXEERIXG MECHANICS. 

Boilers up to 9' diameter may have 1 fiirnace. 
13' 6" " " 2 furnaces. 

" beyond 15' " " 4 

in each head. 

Approximating the Length of One Boiler. — Allow 12 inches in 
length of furnace for the bridge-wall, and allow for furnace door, 
flanging, etc., a distance of 4^ inches from the outside of front tube 
sheet to the commencement of grate. This distance is simply taken 
in order to get a basis of work from some fixed point of the boiler. 
In this case the outside of the front tube sheet is taken. The 
furnace ends will still project be3'ond this a sufl&cient amount to 
allow for riveting the furnace tubes to the head of the boiler. 

Length of Combustion Chambers. — When two furnaces empty 
into the same chamber, this is usually about ^ of the length of the 
grate. 

For this dimension, practice seems to vary within considerable 
limits. (See Seaton, p. 460.) 

Eecent practice in the Bureau of Steam Engineering is to make 
the length of combustion chamber, for furnaces of 40 inches in 
diameter, two furnaces emptying into the same chamber, about 30 
inches, but in any case space enough for a man to work in must be 
allowed. Make water back between opposite combustion chambers 
6 inches at bottom, increasing to 10 inches at top. 

Combustion chamber sheets for boilers working with over 70 
pounds pressure are made of ^ inch plates. 

Then, approximately, the total length of the double-ended boiler 
is 2 [dead plate + furnace + bridge wall -}- combust, chamber + comb. 
Cham. sheet]-F6"r=2[4i-f6'4"4-12"-|-2'6"-f-i"]-f 6" = 21 feet. 

From the above, length of tubes = dead plate, 4^" -f furnace, 6' 4" 
+ bridge wall, 12" = r8r. 

Approximate Diameter of Boiler. — Seaton says good results (for 
forced draught) can be obtained with an allowance of IJ to IJ cubic 
feet of boiler volume per I. H. P. 

40^8 = 5 = Xo. boilers. 
13,500 



5 

13,500 

,5 



X l-o =4050, vohime of one boiler in cubic feet. 
Xlj: = 3375, volume of one boiler in cubic feet. 



NOTES ON THE DESIGN OF CYLINDRICAL BOILEKS. 261 

4050 



21 
3375 



1 92.8 =z area of head in square feet. 



= 160.7 = area of head in square feet. 

For an area of 192.8, diameter = 15' 6". 

For an area of 160.7, diameter = 14' 3". 

So that a diameter of 15 feet 3 inches just inside the limiting 
diameter will give good results: .'.we make the diameter =15' 3". 

The number of furnaces in one head, hj reference to Seaton's 
tables, or by plotting the circle of the head, and roughly putting 
in the furnaces, is found to be four. 

Thickness of shell: 

, factor of safety = 4. 5. 

A=^_;p = 160; f = 58,000 to 62,000 for mild steel; 
^ percentage strength of joint = 85. 

rj,... , 160x183x4.5 _,„^. ^,,, 

Thickness, t= gg^ooo^a^gS =1-2- = ^ ■ 

Board of Supervising Inspector's rule : 

P = pressure in pounds per square inch. 
t= thickness in inches. 
f = strength of material. 

ft 



.80P = 



6xrad 



, .8x160x6x183 , 1 .. . , 

'= 62,000X2 + 16 (fo'- corrosion). 

t=1.14 + TV 

. ' . Thickness of sheets = 1 J". 
Area of Grate: For One Furnace = Q' 4:" x 3' 3'' = 20. 5S sq. ft. 
20.58x8 = 164.6 sq. ft. for one boiler. 

Heating Surface of Furnace. — The ratio of length of corruga- 
tion to length of flue will be about 1.175. 

.'. Heating surface = | X ^i>lMM6 x 7^x1.175 = 370 sq. ft.* 

^ I/O 

* There are 8 furnaces in the boiler, but the upper semi-circumference 
only of the furnace is effective as heating surface, hence ^ . 41" is the 
mean outside diameter of the furnace. 



262 engi?^eeri:ntg mechaxics. 

Heating Surface of Combustion Chamber. — By measuring a 
rough sketch, by means of a planimeter, it is found that the heating 
surface of the combustion chambers at one end is slightly in 
excess of the heating surface of the four furnaces emptying into 
them. This surface for this boiler is estimated at about 400 square 
feet. 

The total heating surface that a cylindrical boiler may contain 
is = [diameter of shell] ^ X [length of tube + 3] for a single-ended 
boiler. For a double-ended boiler this will be changed to [2 x length 
of tube + 3]. From the above rule of Seaton's: 

Total heating surface^ (ISJ)^ x (2 x ^|I + 3) =4981 square feet. 

In modern practice the heating surface varies from 2 to 2.3 
square feet per I. H. P. While considerable care and skill are re- 
quired to keep it as low as 2 feet and still obtain good results, 2.3 is 
of common occurrence. Taking a mean we have 2.15 square feet 
per I. H. P., and for one boiler, 2700x2.15 = 5805 square feet. 

Tube heating 5'wr/ace = 5805 — 370 — 400 = 5135 square feet. 

The area through tubes should be about y the grate area, with 
natural draught (Seaton). In modern boilers designed by the 
Bureau of Steam Engineering the calorimeter through tubes is 
nearer ^. For a mean, take ratio of grate surface to area through 
tubes of 7.5. 

Tube area = 164.6 ^7.5 = 21.94. 

Diameter tubes,- outside = 2^" thickness 12 B. W. G. about '.'12. 

Internal diameter = about 2". 

, ,, . , . xd- 3.1416x4 

Area through one tube= - — -— - = — , — ttv " • 
^ 4x144 4x144 

Total- numher of tubes= ->,>- X -Fr^rT-^n =1008. 
' /.5 3.1416 

IN'umber of tubes in each end = 504. 

Pltcli of tubes, about 1.5d and greater if possible. 1.5d = 3f" 

_ q 1 // 

— 02 . 

This distance, 3-J inches, is the standard practice of the Bureau of 
Steam Engineering for a 2^-inch tube. Pitching these tubes on the 
tube sheet, about 532 can be placed, as shown, in the four nests. 
Allowing ample space between nests and over furnaces, number 
of tubes actually put in each end of a boiler = 532. 

In calculating the tube heating surface the total length of the 
tube was taken and the portion of the tube sheet between the tubes 



NOTES ON THE DESIGN OF CYLINDRICAL BOILERS. 268 

was neglected in calculating the combustion chamber heating 
surface. 

Length of tube = 7' SV, circumference of 2J:" = Tr0686. 

rr 1. XT Q 2x532x92.5x7.0686 ,^^^ 
TubeH.S.= 121<T2 - ^^^^^^• 

Total H. S. = 4831 + 370 + 400 = 5601 square feet. 

Eatio of H. S. to grate surface =^'1^ =33.82. 
^ 164.6 

Laying Down Boiler on Drawing Board. — Draw a transverse 
view of shell and divide it by a vertical center line. One-half of 
this view is to be shown in section (the section taken between 
front head and front combustion chamber sheet) , the other half in 
elevation. From the center, describe the arc of centers of fur- 
naces. Leave 2 inches between the out&r surfaces of furnaces. This 
is practically settled by the fact that the front head sheet must 
be flanged outward to receive the furnaces and a certain distance 
is required for this operation. The furnaces are 45" from center 
to center, measured on chords of this circumference. The two 
middle furnaces have their centers 22^" from the vertical line. 
Since the corrugations of the furnace alternate, one furnace in 
the sectional view will be shown through the bottom of the corru- 
gations 39 inches in diameter, and the adjacent furnace will show 
the maximum diameter 43 inches. Draw the sectional view of the 
combustion chamber with a water space of 4 inches betw^een it and the 
shell at the bottom, increasing to 7 inches near the top of the tube:^. 
Between the two combustion chambers allow a 6-inch water leg. 
Between each nest of tubes over the furnaces allow about 6 inches 
clear for circulation, examination and cleaning. As this boiler will 
require three sheets to make up one circumferential belt, we must 
locate the longitudinal joints. In order to be clear of obstructions, 
let one joint be about 45° with the vertical, and the remaining 
joints 120° from this one: the joints of successive belts breaking 
joints with one another. 

The longitudinal joints will be double-strapped butt joints and 
the details will be found under Eivets and Eiveting. 

The tubes will be of mild steel 2^ inches external diameter, No. 12 
B. W. G. and swelled to 2 ^g- inches at front end. The back end of 
tubes will be expanded into the tube sheet, beaded over into a 
counterbore, which will be tilled with a ring for protecting the 



264 ENGINEERING MECHANICS. 

ends of the tubes from the action of the flame. Any mode of 
protection ma}^ be used, subject to the following conditions : 

1. That it will not interfere with the use of ferrules. 

2. Will not cause injury to the tube sheets when the tubes are 
cut ont. 

3. Will not rednce the area through tubes. 

Stay Tulyes. —The^e will be "^o. 6 B. W. G. thick and will be 
reinforced at both ends to an external diameter of 2f inches, the 
bore of the tubes being uniform from end to end. The}^ will then 
be swelled out at the front end to 2^ inches. The stay tubes are 
threaded parallel at the combustion chamber end and taper at the 
front end to fit threads in the tube sheets. They will be screwed 
into the front tube sheet to a tight joint and will be made tight at 
the back end by expanding and beading. 

The holes for stay tubes are tapped at one 
operation by a pair of taps on a rod as sho^n in 
Fig 122a ^-^^^ ^^^^^ ^^^ ^ duplicate of this is always carried 

among the spare machinery of the ship. 

Cast-iron ferrules will be used to protect the ends of the stay 
tubes at the combustion chamber ends. The space between tubes 
should be sufficient to permit free circulation of water and allow 
the steam to rise freely to the surface. The tubes should be 
aranged in horizontal or nearly horizontal rows, i. e., they should 
not be zigzag. They should not be crowded, as the steaming 
capacity of the boiler may be seriously impaired thereby. 

The top of the top row of tubes should not be placed nearer 

to the top of the boiler than at a distance equal to one-third of the 

internal diameter of the boiler. 

15' 3"_2"5 

==5' to top row of tubes. 

o 

From this the center line of the top row of tubes can be 
obtained and the tubes pitched horizontally and vertically. In 
laying out the tubes, begin at the lower row, leaving 6 inches clear 
above the furnace for wing furnaces and 13 inches for the bottom 
furnaces. This is done for practical considerations and to give free 
outlet for the bubbles of steam forward on the crown sheet. The 
top and inboard side of the combustion chamber can then be com- 
pleted. 

Having put in the furnaces, combustion chambers and shell, 
in the transverse view, lay off the divisions into which the head 



XOTES OX THE DESIGN OF CYLIXDEICAL BOILERS. 265 

sheet is divided. The upper sheet is cut with an allowance for 
lap sufficient for a double riveted lap joint (zigzag) leaving just 
sufficient metal clear above the tubes to permit of expanding the 
tubes (-J- inch to 1 inch will be sufficient) . La}^ out the clear dimen- 
sions of the tube sheet for similar riveting at top and bottom. 

In cutting the lower part of the front tube sheet let it enclose 
the manholes to be placed in the wing spandrels and above the 
middle furnaces, and thus act as a strengthening ring. The lap 
around these manholes should be sufficient for single riveting. In 
cutting the tube sheet from the wing to the middle nest of tubes, 
let the slope be about 30°. 

The manholes in the Aving spandrels will be three-sided, about 
13"xl5", with the corners rounded and the sides curved to con- 
form to the curves of the shell and furnace. The center man- 
holes are to be elliptical, 11" x 15". Between the center furnaces 
and nests of tubes, support the flat head by braces about 10 inches 
apart, three over each furnace and one at the corner of the wing 
manholes. These braces tie the head sheet to the front plates of the 
combustion chambers. In the lower spandrels cut an ll"xl5" 
manhole; and at each side of these manholes and at the top 
strengtlien the head by a brace ; the upper one to continue through 
to the opposite end of boiler or to the combustion chamber; the 
lower ones to be diagonal and connected to the shell. 

The back tube plates vary in thickness from y-g- inch to f inch, 
Seaton. They are generally made, however, from f inch to f inch, 
depending on the size of boiler, as with this thickness the tubes can 
be made quite tight, and there is less liability of cracking the plates 
or burning the tube ends than with thicker plates. 

It is not usual to make the sides and bottom of the combus- 
tion chamber exceed -J in^h in thickness. They are stayed by screwed 
stays and the bottoms are stiffened by angle irons, about 3^" x 
SVxV', riveted to them. The screw stays are pitched to suit 
this thickness of plate by the method given in Unwin, viz : 

f=-|-^xp;f = 6000. 

The increased pressures have led to the adoption of corrugated, 
ribbed and spiral furnace flues for additional strength against 
collapse without abnormally increasing the thickness of the plate. 
There is a great objection among engineers to furnaces over 4- inch 
or f inch thick, on account of the increased resistance to the trans- 



266 



ENGINEERING MECHANICS. 



mission of heat through the plate and the consequent overheating. 
While the resistance to the transmission in a laminated material 
may be abnormally increased by the thickness, yet in a homo- 
geneous metal like steel it would seem that the thickness ordi- 
narily used might with safety be increased, for, from experiment, 
the condition of the surfaces seemed to have a much greater value 
than the internal conductivity. 

From the measured rate of transmission along bars and plates, 
when fairly clean and free from laminations, it was found that a 
furnace plate might be increased from ^ inch to f inch without 
increasing the resistance more than 1.25^. 

In general, the practical thickness used for 160 pounds pressure 
is i inch and f inch. 

The following is the formula generally adopted for corrugated 
flues : 

CXt 



t = thickness of plate in inches, 
p — working pressure in pounds per square inch. 
d = mean diameter of furnaces in inches. 
C== constant having values from 12,500 to 15,000. 
For Morison furnaces a value of 15,000 is used. 




Fig. 123. — Morison Suspension Furnace. Size of Corrugation. 



This value of C will depend upon the kind of furnace and of 
course upon the strength of the metal used. These furnaces are 
always made of steel of about 27 tons ultimate strength. For 
an intelligent assignment of a value to C, it is safest to obtain 
the latest edition of the maker's circular or to consult the value of 
C as assigned by authorized rules of inspection. 

In the case of the boilers in hand, 

160X41 



t: 



12,500 



1 6 • 



NOTES ON THE DESIGN OF CYEI.NDlflCAL BOIi.EKS. 



267 



e-e 



Riveted Joints. — In general^ for thin plates, where two plates of 
unequal thickness como together, it is customary to proportion 
the joints to the thinner plate. This rule will not always answer 
for thick plates, and to secure tightness at the joint 
it is proportioned, if not for the thinner plate, for a 
mean proportional between the two. 

For the method of obtaining the formula used in 
proportioning joints see notes on riveting, Art. 134. 
They are mainly from Trail and Foley and are given 
as representing the best modern practice and are appli- 
cable to thick and thin plates. 

For the connection of hack end of furnaces to combustion cham- 
ber, and for the joints of adjacent ring furnaces, use a single riveted 
lap joint. See Fig. 124. 

t = r; ^=54.73. 



Fig. 124. 



, 1.55 X 54.73 X j ^//93^^15/' 
45.27X1X1 '' 



100 (p-d) . 
P 

Lap sheet; 



1 5« 



54.73. .•.45.27p = 100xl| 
2E = 2xl.5dr=3d = 3xlf = 



'T6" 



For the Upper Head Sheet and Tube Sheet. — Tube sheet =-J'', 
upper head sheet=:lj". Making rivet section and plate of equal 
strength, fc = fc^, and d = l''25 = thickness of sheet. 

The joint is double riveted, lap (zigzag). See Fig. 125. 

p= ?3xAxnx.c 4_ (:[^ 2.308 + 1.25 = 3.558 = 3yV'- 

/v o X t 

In the above, t is taken as J, that is, the rivet is 
proportioned to the thinner sheet. 



Q 9 



Strength of joint— ^ 

Eivet section, <, — 64.7. 



^ = 64.9^. 



E 



2" A ±4 — Is • 







I5 





Fig. 125. 



.._ V(llp + 4d)(p + 4d) _ V(39A+5)(3tV+5) _ 

10 10 -1-.'4D01 ly^. 



V=lif". Lap of sheets = 2P] + Y = 5|i". 



268 EXGIXEEKIXG MECHANICS. 

Connection of loiver edge of tube sheet to head sheet: double 
riveted, zigzag, lap, joint. 

d=:t=r. 

p^ 23xA2<nxc 4_clxl.l29 + .875=:2:'004. 

The percentage strength of plate and rivet sections are: 

fr = 56.56 (plate), and ^^ = 56.U (rivet). 

If the lower head sheet had been taken at f inch and the joint 
made for equal strength of plate and rivet sections, the joint would 
more nearly have approached that of the upper one. For example, 
let t=f" (thickness of lower head), 

and the strength of plate = [-^^\—^] 100 = 61.1;^, while the strength 

. . , 100 X 23 X 2 X. 60132 ^^. 
of nvets= g8x2ixf "^^^^ 

This latter proportion for the joint would probably be better, 
hence the lower plate could be made f inch and a very fair joint 
made. This sheet being thinner would also make the flanging for 
the furnaces easier. 

Longitudinal Shell Joints. — Butt joints, double strapped, treble 
riveted with alternative rivets omitted in the outer row. See Fig. 126. 

Percentage strength of joint = 85. 



>-Q-Q-(^-0-^ 



©■■©■ 0~Q--^-i 
-©-©-■©-©■-©- 
I© Q -e- 



-,_ 1.55x85xt . x_-,i./ 
'^"15X5X1.75^^--^^- 



., , 1.55x85x1.25x100 q„^^ 

* P'^°'^-P== 15X15X5X1.75 =^-^^- 

orP=i' = .85. 



Fig. 126. 



P 

15p = lJ. .-. p = f X-V/=8r33. 

p = 8f ", greatest pitch or pitch of outer 
row. 

Middle Circumferential Joint. — ^Treble riveted, lap, zigzag. 
t = ir, d^lf". See Fig. 127. 
^^23X1.484X3X1 +i|^2.93 + 1.375 = 4.305 = 4yV'. 

/io X -L. vO 



NOTES OX THE DESIGN OF CYLIXDRICAL BOILERS. 



269 



y^ V(47A + 5i)(4A + 5i) _oog. 



10 



'T6" • 



E=fd:=2J, 







©?--^ 


^' 


-^(>-- 


w 


<^-(^ 






1 




1 





Lapof sheet=2iV+23%+23V+2 \=Sf\ 
Strength of above joint. 

^ = /^3tH^ 100 = 68.1, plate. 

Circumferential Seams at Boiler Ends. — Owing to the additional 
stiffness imparted by the heads of the boiler, these seams are 
double riveted instead of treble riveted as in the middle seams. 
The joint is double riveted, zigzag. In this case as the shell is IJ" 
and the tube sheet f inch the conditions are the same as for the joint 
between the tube plate and the head sheet, and the same proportions 
will be used, p = 3 -^Y ; d = li" ; lap = 5^" ; V = lif" • 

Width of Butt Straps, Distance between Rows of Rivets, etc. — 
From center of rivet to edge of sheet=f d = lf". 



— /I 3 // 

6 • 



Y= — ^ — ^ ^ ^P — \ in which p is ^ greatest pitch =4^ 
Y- V(46TV+5 y(9A)-o 168-9 a " 

Yi = — ^^ — P ^ ^P '- where p is the pitch of the outer 



row or 



y^^ V(92i + 25)(8f + 25) ^3 -^g^g_3_.,^ 

V/idth of butt strap = 4E + 2y + 2Vi = 7i + 4f + 6f = 18i. 
8.375-1.25 



Plate section : 



8.375 



85.37^. 



-D- . X- 23x1.227x5x1.75x100 o. or-^ 
Eivet sections ,3 ^ i.,5^8.3,5 =84.26^. 

Thickness of Butt Strap. — T = tx 



p-d 



)= 



95 or 1" for each 



p-2d 
strap. 

Longitudinal Section of Boiler. — The length of the boiler is, 
from the details as fixed in the preceding work, 20 feet 10 inches, 
and is made up of three rings. The end seams are double riveted 
18 



270 



ENGINEERING MECHANICS. 



and the raiddle seams treble riveted. The distance from the plane of 
the front tube sheet to the edge of the shell is 4 inches. The lap of 
the middle seams, previously calculated, is 8f inches. Hence, the 

1 4.-U 4. • 20'10" = 2x8f"-2x4" 21' 7^" ^. ^,„ 
length of one ring= — ^ = ^ — =7 2^ . 

o o 

Draw the longitudinal view of the shell, cutting the sheets as 
shown in Fig. 128. 

The upper ends are to be rounded to avoid extensive bracing. 
Leave a sufficient amount of flat space immediately above the upper 
joint of the head sheet for one row of braces with their washers, 
and then round the sheet to an arc of a circle (quadrant) : a radius 
of about 3 feet 7 inches will effect this. 




Fig. 128. 



N"oTE. — Curved ends of C3dindrical boilers : 
R = Eadius of curved end in inches. 
T=: thickness of tube plate in inches. 
d = outside diameter of tubes. 
p = pitch of tubes. 

r = ratio of solid metal left between tubes to pitch of tubes, 
i. e., 
d 



r = 



P: 

Then R 



boiler pressure. 
CxrxT 



, where C = 13,000. 



If the percentage strength of the horizontal seam above the 
tubes be less than the percentage strength of plate left between 
the tubes, such lesser percentage should be used in calculating R. 
If the plate above the tube plate be less in thickness than the tube 
plate, its thickness should be used in the formula; that is, the 
smallest of the products r X T should be used. 

Put in the head sheets and make all the joints of the shell. 
Project the furnace and combustion chambers and lay off the 
corrugations and laps of the furnace connections. Project the 



XOTES OX THE DESIGN OF CYLIXDIUCAL BOILERS. 271 

tubes across and put in all screw stays both in the longitudinal and 
transverse views. 

Curved Tops of Combustion Cliamhers. — The combustion cham- 
bers are to have rounded tops, opposite ones to be tied together by 
gusset stays, and the rounded parts to be further stiffened by angle 
irons. 

On account of the want of continuity, the curved tops of the com- 
bustion chambers are not capable of standing the same pressure 
as a complete cylinder of the same radius, and should be stiffened 
by angle or T bars. The constants for plain cylindrical furnaces 
should be considerably reduced when applied to curved tops, as 
the tops have been known to come down at about one-half the 
pressure suitable for a round furnace of the same plate thickness 
and radius. The curved tops should be efficiently stayed to the 
ends of the boilers, or with double combustion chambers back to 
back they should be efficiently tied together. The breadth of the 
combustion chamber in inches multiplied by the radius of curva- 
ture in inches is the least surface for which stay power is required. 
The radius of curvature should not be less than the width of the 
chamber, so as to avoid fiat surfaces on the top and near the tube 
sheet. The bottoms of the combustion chambers are also stiffened 
with angle irons. 

Foi' determining the size of girders vjliicli support the flat top of 
the comMistion chamber, the girder is treated as a beam supported 
at the ends and loaded at three points, as usually each girder carries 
three stay bolts. The formula for this is from Church's Mechanic's, 
page 2 TO, and is : 

M = fZ= ^^^=r Pol, -P,(L-1,). where, 

f = 8500 pounds safe stress in outer fiber of girder. 

I = moment of inertia of section= --— . 

y=~(^ , distance of outer fiber from neutral axis. 

P^=:load at one end of girder. 
P^ = load on one girder stay bolt. 

1^=3 one pitch of girder stay bolts. 

l^=:two pitches of girder stay bolts. 

b = breadth of girder = If" in this case. 

d = depth of girder. 



^'J'^ EN-GINEteRIXG MECHANICS. 

Length of girder equals depth of combustion chamber. 
The girders are arranged in pairs, so b is the breadth of the pair. 
Head Braces. — The area supported by each brace by measurement 
on the drawings is 15" X 14.5". 

. 15x14.5x160 oQ^ • -u 

A=: — — =3.87 square inches. 

d = 2r25 at bottom of thread. 
Diameter of washer =6". 
Thickness of washer = f ". 

Braces over Center Furnace. — Area supported by each about 
15"X10". 

Areabracezr ^^ ^^^^ ^^^ =2.67 square inches, 
y uuu 

.-.d^if". 

N'uts, threads, etc., from P. I. 

8tays helow Furnaces. — The positions of these braces are fixed by 
the relative positions of the shell, furnace flues and manholes. 
Having fixed their positions as at XYZ, Fig. 129, the braces 

may be proportioned by supposing 

them to support the triangle (XYZ) 

) fi, \ 1 formed by joining them, together 

^/;^^\ J with half the area included between 







io^±'-\z±^--^^^f 



^'V^'ir^o. y^ the sides of this triangle and the 

adjacent shell and fines. To ap- 
proximate to this area, erect per- 

FiG. ] 29 ^ r 

pendiculars at the middle points of 
the upper sides of the triangle. Take B, C and E half way between 
the sides of the triangle and the circumference of fines, etc. Draw 
ABG, G-EF and ACF, so that these lines divide equally the area 
between the sides of the triangle, XYZ, and the fines or shell. The 
braces are calculated so that X supports ABDC, Y supports CDEF 
and Z supports BDEG. 

For the present case : 

Area supported by brace X = 128 square inches. 

Area of brace = — — — - — =2.27 square inches. 

.*. diameters 1'.' 7. 
Area supported by stays Y and Z = 120 square inches. 
Area of stay = 2.13 square inches. .*. diameter = If". 



NOTES ON THE DESIGN OF CYLINDKICAL BOILERS. 373 

Of these three stays the ujDper one will be a through stay pass- 
ing between the combustion chambers to the opposite head; the 
lower stays will be diagonal and secured to the sides of the shell by 
" palm ends " and rivets. 

Diagonal stays are never used if a horizontal stay can be used, 
as the resultant tension is greater on the diagonal stay. 

If a = area of diagonal stay, 

_ area supported in square inches X pressure . 

Where A is the angle the diagonal stay makes with the bottom 
of the boiler, in the same radial plane as the stay. 

The braces around other manholes are direct stays fastened to 
the head and to the front sheets of the combustion chambers. 

The washers used for main stays are riveted to the head. 

Screw Stays. — The combustion chambers are stayed to each other 
and to the shell of the boiler by stays screwed into both sheets and 
fitted with nuts ; the nuts to be set up on beveled washers where the 
stays do not come square with the plates. The holes for screw stays 
will be tapped in both sheets while they are in place. 

All screw stays and braces will have raised threads ; all braces will 
be made without welds. 

Pitch of stays = 7". 

Area of stav = ^^J^^^^^^ =1.306 square inches. 
dOOO 

Diameter =1'.'25 or IJ over threads. 
For Pitcli of Stays and Braces. — The pitch, 1, of the stay bolts 
and braces depends in a measure on the thickness of the flat plates 
supported, as will appear by treating the square surface supported 
as a beam fixed at the ends and uniformly loaded. The formula for 
this is : 

fXI WXl , , 65,000 T IXt^ t ^, ^,„ 

— - = -12- ^ where f = -^-^ , 1= -^^ , j= -^ , W = pXP. 

The area supported by each stay or brace is P. If p is the pressure 
per square inch on the flat surface, then the total pressure supported 
by each stay or bolt will equal p x P. The safe load per square inch 
of section of good steel stays and braces is 8000 pounds, therefore 
the area of each stay must be such that : 

8000x-^=pxP. 



274 EXGINEERIXG MECHANICS. 

The diameter d of the stays is taken at the root of the threads, 
where the ends are fitted with nuts. 

Grate hars are to be of wrought iron, cast iron or steeL 

Bridge lualls are of cast iron; to be readily removable, finished 
with fire brick. 

Furnace fronts are to be made with double walls of wrought iron, 
the space between the walls to be in communication with the ash 
pits. The upper part of inner plate is to be perforated as directed ; 
dead-plate to be of cast iron and easily removable. 

Furnace doors must be protected in an approved manner from 
the heat of the fire, and open inward. 

Ash-pit doors to be made of -J" iron, stiffened with angle irons. 
False ash pans of ^-inch iron are to be fitted under all grates. 

Circulating plates are to be fitted at each side of each nest of 
tubes and made so as to be readily removed and introduced through 
manholes. 

The uptahes are to be of wrought iron, K'o. 8 B. W. G., built on 
channel irons, stiffened with angle irons, and bolted to the boiler 
heads and shells. Outside of the uptake is to be a jacket enclosing 
a 3-inch space. This space is to be filled with an approved non- 
conducting material. 

Uptake doors are to be double shells of iron, TTo. 8 B. W. Gr. 

The space between the shells is to be fitted with the same non- 
conducting material as used for the uptakes'. 

Smo'ke pipes are to be two in number; the top about 85 feet above 
the grates, the bottom part shaped to join the uptakes. The top is 
to be finished with angle irons and with a hood covering the casing. 
The pipes are to be stayed by shackles and guys with turnbuckles for 
setting up the guys. 

A casing enclosing a 3-inch air space extends from the uptake to 
about 6 inches below the hood at the top. An additional casing is 
placed for 5 feet or more above smoke pipe hatches. 

Boiler attachments are to be supplied as follows : 

1 self-closing main steam stop valve. 

1 self-closing auxiliary steam stop valve. 

1 dry pipe. 

1 main feed check valve with internal pipe. 

1 auxiliary feed check valve with internal pipe. 

1 bottom blow valve with internal pipe. 

1 surface blow valve with internal pipe and scum pan. 



NOTES OX THE DESIGN" OF CYLINDEICAL BOILERS. 



275 



2 safety valves to be connected with dry pipes, or to have internal 
pipes. 

1 steam gauge, on single-ended ; 2 on donble-ended boiler. 

2 glass gauges with automatic cocks on single-ended, and 3 on 
double-ended boiler — two of them being at feeding end. 

4 gauge cocks on each end of double-ended boiler. 

1 sentinel valve of ^-inch area. 

1 salinometer pot. 

I drain cock. 

1 air cock. 

1 approved circulating apparatus. 

1 cock threaded for the attachment of a syringe. 

All external fittings, unless specially stated to the contrary, are 
to be of composition. 

All fittings are to be flanged and through bolted. 

All cocks, valves and pipes are to have spigots or nipples passing 
through the boiler plates. 

Internal pipes of brass must not touch the boiler plates except 
where they join their external fittings. 

The stems of all valves on the boilers are to have outside screw 
threads. 

Securing Boilers in Iron and Steel SJiips. — The boilers are gen- 
erally placed on board the ship after launching, as the vessel being 
waterborne is less liable to be strained from the concentration of 




Boiler Saddle, Port Side, Looking fbrwairi, Jog at "P" 

for Butt-Strap. 



Section N-0. 



Fig. 130. 



weight. The framing of the boiler hatches should be so arranged 
as to give ample room for lowering the boilers into their com- 
partments. The ship is taken under the shears or derrick, the 
boilers, slung by straps, lifted on board and lowered on blocking. 
They are then moved along the floors by jacks until in their proper 
positions over the saddles and are then lowered into place. The 
boilers rest on saddles and are secured to them bv bolts. The 



276 



ENGINEERING MECHANICS. 



number of saddles depends on the size and kind of boilers used, 
being generally two for single-ended and four or five for double- 
ended boilers. They should be spaced so as to allow access to the 
circumferential seams for caulking. Where double-ended boilers 



A 



iji-i-^4i4n 



" "^ 'Angle 



K 



Chock 



Showinoj Saddles in Position. 
Fig. 131. 

are used, the fire rooms are generally athwartships. The saddles 
are then arranged so as to come directly over the frames. Each 
saddle consists of a vertical plate secured to the inner bottom and 
reverse frames by angles. The top of this plate is cut to the cur- 
vature of the boiler, to which it is secured by angles on each 



H 



Stiff Hed Lead 
(Joint- 
Boiler Shelly 




Bolt to be a snu^ fit m Boiler Shell. 
Steel Wasiiers. 



fBolt iiole in ttiisAn^le Iron 
Aflliptical with tAe longest 
(Diameter Fore and Aft. 



Fig. 132. 



side, jogged if necessary to fit over the longitudinal butt straps. 
The boiler rests on the angles. The boilers are screwed to the 
saddles by bolts and nuts. The bolts are screwed through the 
shell from the inside and have a cup washer under the head, the 
joint being made tight by red lead putty under the washers. The 
holes through the flanges of the saddles are elongated in a fore 



NOTES ON THE DESIGN OF CYLINDRICAL BOILERS. 277 

and aft direction to allow for expansion. In addition to the above, 
some means must be adopted to prevent displacement in a fore 
and aft direction, if the vessel is intended to be used for ramming. 
The usual method is to fit what are called ramming chocks. These 
consist of a vertical plate about 1 inch thick fitted to each end 
of the boiler and secured to the inner bottom by angles. They are 
placed opposite the lowest point of the boiler and are fitted so as 
to extend about 6 inches above the lowest point and a clearance 
of about one-quarter inch is left between them and the boiler. In 
large boilers two are used at each end. When the boilers are placed 
with their axes athwartships the saddles should be placed over 
the longitudinals; when this is not possible, heavier plates should 
be worked in the inner bottom below the saddles. The saddles 
and the methods of securing the boilers to them are illustrated in 
Figs. 130, 131 and 132. 

134. Riveting.— FormulcB for Proportioning Riveted Joints and 
Calculating the Width and Thickness of Butt Straps for Steel 
Boilers. — With the aid of these formulae the tables for riveting 
joints in TrailFs work are calculated, but for every-day work in the 
designing room it will undoubtedly be a time-saving operation to 
obtain the desired information direct from those tables. For con- 
venience in comparing with the above-mentioned tables the same 
nomenclature is used. The following assumptions are made : 

First. That the mean tensile strength of the plate is 28 tons 
and the mean shearing strength of the rivets is 23 tons per square 
inch of net section. 

Second. That rivets in double shear offer 1.75 times the resist- 
ance to shearing exerted by rivets in single shear. 
Let p = pitch of rivets in inches (greatest). 
d = diameter of rivets in inches. 
c = constant whose value is 1 for lap or single butt strap, 

and 1.75 for double butt-strapped joints. 
A = area of one rivet in square inches. 
n= number of rivets in the greatest pitch. 
^= percentage of plate left between rivets in the greatest 

pitch. 
^1 = percentage of the rivet section compared with that of the 

solid plate. 
^2 = percentage of the combined plate and rivet section when 
alternate rivets are omitted in the row 



278 EN"GIXEEEING MECHANICS. 

p^ = the diagonal pitch. 

E = the distance from the center of the rivets to the edge of 

the plate. 
Yrzthe distance between the rows of rivets in ordinary zigzag 

riveting, and in chain riveting when alternate rivets 

are omitted in the outer row. 
Vi = the distance between the onter and the next row of rivets 

in zigzag riveting when alternate rivets are omitted 

in the outer row. 
t=:the thickness of the plate in inches. 
ti = thickness of a single butt strap in inches. 
t2 = the thickness of a double butt strap in inches. 

To Find the Percentage Strength of any Given Joint. — The metal 
left between the rivet holes = (p — d)t^ and the percentage strength 
of this as compared with the solid plate is : 

M(£:zd)xt. .•.^=M(P^. (1) 

pt P 

The resistance offered to shearing by the rivets in a single pitch 
is 23xAxnxc; comparing this with the strength of the solid 
plate = 28 X p X t, we have : 

cf — lQQx23xAxnxc ,„x 

28xpxt • ^^^ 

Considering now the case where alternate rivets are omitted in 
the outer row, in the next row there will remain (p — 2d) xt solid 

metal^ and ~ — ^^ will be the percentage strength of this row 

P 

as compared with the solid plate ; but^ supposing that the plate tears 

along this row of rivets (see Fig. 134), before a total giving 

away of the joint occurs one rivet for each pitch must also be sheared 

in the outer row and the percentage strength of this single rivet is 

— ^ , which value will be obtained from equation (2). From the 

above considerations we have 

.^ ^ lQ0(p-2d) i, .3. 

' ^ p n 

The lowest of the values obtained from equations (1), (2) and 
(3) is the percentage strength of the joint. An examination of 
these equations shows that for double butt-strapped joints, so 



IsTOTES OX TPIE DESIGN OF CYLINDRICAL BOILEES. 279 

long as the diameter of the rivets is not less than the thickness 
of the plates, '/c^ is always greater than ^i^ or </c-^. This is also the 
case with lap joints, so long as the diameter is not less than 

^^ o ., . . ^ = ^TT^^T^ , and, as both these conditions nsually hold, 

23 3.1416 .64515' ' ^ ' 

28 ^ 4 

the TTse of formula (3) will seldom be necessary. 

To find d, when p, c, n and t are given, so that ^ may equal fc^ 
(that is, so that plate and rivet section may be equivalent in 
strength). 

From (1) and (2) we have: 

lOO(p-d) ^ 10Qx23xAxnxc ... 

d 28xpxt • '^ ^ 

Substituting for A its value =—^ — and simplifying, we 

have : 



-]2i 1.55td ^ 1.55pt 
nXc nXc 



Solving for d we have : 



-, _ //^oty , 1.55pt _ .775t 
VUxcj "^ nXc nXc ' 



which reduces to : 



//775t/.775t , _ \ .775t /^x 



If d, c, n and t are given; from equation (4) we have: 

23xAxnxc -, , . 

To find d and p, when n, c and t and ^=^i are known : 
From (1) we have 

lOOd ,„. 

22d2 
Substituting this value for p and letting A— ^y . in (2), we 

have: 

100x23x22xd2xnxc 



i^ir^ 



^„ ^ , lOOd 
^8X7X4X j(3^3^Xt 



280 ENGINEERING MECHANICS. 

Solving, 



-,_ 1.55x^Xt ,^. 

^~ (lOO-^)xiixc* ^ ^ 

Substituting this value of d in (7) we have : 

lQ0xl.55x^Xt ,Qx 

^~ {100-fo)'XTLXc' ^ ^ 

It will generally be found simpler to substitute the value for 
d found from equation (8) in equation (7), thus obtaining a value 
for p direct, without using equation (9). 

Diagonal Pitches and the Width of Butt Straps. — The value 
of the metal in the diagonal pitch p^ is only about f of what it is in 
the horizontal pitch, or in that part of the horizontal pitch to which 
it is required to be equivalent in strength. In any case, the diagonal 
pitch should not be less than that found by the following formulae : 

I. 

Ordinary zigzag riveting and chain riveting with alternate rivets 
omitted in the outer row. 
f — F^ V Eeference to Fis^. 133 shows that the same reason- 
<"*— -5^^ ijig applies to both cases, as in each case the net 



:c-?fe: 



■^ section of two diagonal pitches must be made equiva- 
FiG. 133. lent in strength to the net section contained in the 
greatest horizontal pitch. The liability of the sheet to tear along 
AB and BD would be the same as along AD or BF. 

Now the metal left along ABD is 2(pd — d)t and along AD or 
BF is (p-d)t. 

.•.2(p.-d)t:=f(p-d)t,andp.= ^P±^. (10) 

From the right angle ABC we have for the distance between the 
rows of rivets : 

o^y^ V(llp + 4d)(p + 4d) (11) 

The authors mentioned before, state that, for chain riveting, 

this distance should not be less than — ^ — and, as this result is 

greater than that obtained from (11), it is the one that is usually 
found tabulated. 



NOTES ON" THE DESIGN" OF CYLINDRICAL BOILERS. 281 

The distance E from center of rivet to edge of plate should 
not be less than 1.5d; so that the minimum lap of sheets, if lap 
joint, or half width of butt strap = 2E + V. (12) 

II. 

Zigzag riveting with alternate rivets 
omitted in the outer row. /. a p ^ ^^ 



An examination of Fig. 134 shows that ll^^T^. ^.^.q-.^ 
rupture may occur by the plate giving away ^^ 8— c - 



along ABCD^ or along AD, and to make the fig. 134. 

joint symmetrical in strength, the net sec- 
tion along ABCD must be equivalent in strength to the net section 

along AD. Now BC = -^- and the net section along BC is f ^ — d J t, 

while on AD it is (p — d)t; hence the section to which the metal 
left on the lines AB and CD must be equivalent in strength is: 

(p-d)t-[-|- -d)t= ^t, but the net section on AB or CD is 

(p,-d)t; therefore, 2(p,-d)t=|(^-|-tJ , and 

3p + 10d /.o\ 

In the triangle AFE we have FE = — ^ = -^, and, for the 
distance between the rows of rivets, we have : 



Tr _ //3p + 10dV f VY _ /9pM-60pd + 100d2 V^ 
^i~ V 1 A " X ~ V ^100 16 



10 I \4. J ~ y 100 16 

/ llp- + 240pd + 400d" V(llp + 20d ) ( p + 20d) 



(14) 



400 20 

As before, the half breadth of butt strap = 2E + Vi. (15) 

As riveting in all ordinary cases is either of the form shown in 
Fig. 133 or Fig. 134, or a combination of those forms, equations 
(10), (11), (12), (13) and (14) have a general application. For 
instance, consider the case of a double butt-strapped, treble riveted, 
zigzag joint with alternate rivets omitted in the outer row; the 
distance V^ between the outer and second row is obtained from 
equation (14), the distance V is obtained from equation (11). In 
this case the half breadth of butt strap =:2E + Yi + V. (16) 



282 EA^GINEERING MECHANICS. 

Thichness of Butt Straps. — From considerations of strength and 
to secure tight joints the aggregate thickness of the butt straps 
should alwa3'S be more than that of the plate. For single butt 
straps ti=|-t; for double butt straps t2 = ft, are arbitrarily taken as 
the minimum values allowed. 

When alternate rivets are omitted in the outer row, the thickness 
obtained above mnst be increased. 

The ratio of this increase is expressed bv-^- — ^rv, which is the 

^ • p — 2d 

ratio of the net section left in the onter row to the net section left 

in the next row. With this form of riveting, the minimum values 

of ti and to will be found by the following equations : 

t _ 9 ( p-d \ ^ _J p-d \ 

BOILER PROBLEM. 

WORKING SHEET. 
Required: A set of boilers to supply steam of lbs. pressure 

to engines, such as are used in a modern naval 

vessel of the class, when developing under an air 

pressure of inches of water, a maximum of I. H. P. 

for consecutive hours. 

Additional specifications 



BOiLEK proble:\i. 283 



1. Name of ship 

2. Maximum I. H. P 

3. Boiler pressure, pounds iier gauge 

4. Air pressure, in inches of water . . . 

5. Coal burned per sq. ft. of grate, assumed from trials. 

6. Coal burned per I. H. P., assumed from trials 

7. I. H. P. per sq, ft. of grate, assumed 

8. Pounds of water per I. H. P. per hour, assumed 

9. Water evaporated per pound of coal, assumed 



284 ENGTXEEKIATG MECHANICS. 

RESULTS OF PRELIMINARY CALCULATIONS. 

10. Grate surface, from (2) and (7) 

11. Grate surface from (2), (5) and (6) 

12. Grate surface from (2), (5), (7) and (8) 

13. Grate surface used 

14. Trial diameter of inside of furnace 

15. Trial length of grate 

16. Number of furnaces, from (13), (14) and (15) 

17. Number of furnaces used 

18. Type of boiler as determined from tracing of boiler space in ship 



19. Limiting diameter of boiler as determined from tracing. 

20. Number of furnaces in boiler head 

21. Length of bridge wall, assumed 

22. Length of hanging bridge wall 

23. Beginning of grate from outside of front tube-sheet 

24. Length of combustion chamber 

25. Distance between backs of same, to p bottom 

26. Approximate length of boiler 

27. Length of tubes 

28. Length of tubes for low boiler, assumed 

29. Volume of boiler per I. H. P., assumed 

30. Diameter of boiler 

31. Numbers of boilers and class 

32. Thickness of shell, Bureau method 

33. Thickness of shell, Board of Supervising Inspectors' rule. 

34. Thickness of shell, Lloyd's rules 



BOILEU PROBLEM. 285 

35. Thickness of shell used 

36. Ratio of corrugated length of flue to axial length 

37. Heating surface of furnace 

38. Heating surface of combustion chamber, from planimeter meas- 

urements of rough drawing of combustion chamber 

39. Total heating surface. Bureau method 

40. Total heating surface, Seaton 

41. Total heating surface method 

42. Total heating surface from above methods 

43. Tube heating surface 

44. Calorimeter 

45. Number of tubes in each boiler 

46. Number of tubes in each end, by calculation 

47. Number of tubes in each end, from rough drawing 



Having completed all of the foregoing preliminary calculations, 
make a rough drawing, to as large a scale as the paper will admit, 
determine all the details, and tabulate the results in following columns 
before commencing the finished drawing on the board. 



FINAL DATA OF ONE ENDED BOILER, ASSEMBLED TO 

COMMENCE DRAWING. 

48. Diameter of boiler 

49. Radius of arc of furnace centers 

50. Distance between furnace centers measured on chords 

51. Horizontal distance of middle furnace centers from center of head 

of boiler 

52. Number of furnaces in head 



53. Inside diameter of furnace. 
19 



286 ENGINEERING MECHANICS. 



54. Outside diameter of furnace 

55. Thickness of furnace 

56. Thickness of shell plates 

57. Thickness of front tube sheet , 

58. Thickness of back tube sheet 

59. Thickness of top head sheet 

60. Thickness of lower head or furnace sheet 

61. Thickness of combustion chamber sheets 

62. Thickness of butt straps 

63. Distance of top of top row of tubes from top of boiler .... 

64. Water space between shell and combustion chamber, at top. 

, at bottom 

65. Water space between shell and nearest furnace 

66. Water space between two adjacent combustion chambers . . 
"67. Water space, minimum, between nests of tubes 

68. Number of combustion chambers in each end , 

69. Tubes plain, number in one end , 

70. Tubes stay, No. .... .B, W. G., outside diameter 

71. Tubes stay, number in one end 

72. Tube spacing, all, horizontally , vertically 

73. Total number of tubes in one end 

74. Length between tube sheets 



RIVETED JOINTS. 

75. Back end of furnaces to combustion chambers, and to adjacent 

furnaces , diameter of rivets , 

pitch , lap of joint 

76. Upper bead sheet to front tube sheet 

diameter of rivets pitch , lap 



BOILER PROBLEM. 



287 



lap 



lap. 



77. Lower edge of tube sheet to furnace sheet 

diameter of rivets , pitch 

78. Middle circumferential joint 

diameter of rivets , pitch 

79. End circumferential seams 

diameter of rivets , pitch , lap 

80. Butt strap , width , thickness. . 

81. Butt joint, kind of joint 

diameter of rivets , pitch , lap . . . 

distance between rows of rivets 

82. Length of boiler 

83. Length of one ring 

84. Radius of curved head 

85. Braces above tubes, diameter 

86. Braces to back tube sheet, diameter 

87. Braces above lower manholes, diameter 

88. Braces, diagonal, in lower water space 

89. Screw stays, diameter , pitch 

90. Heating surface, plate, square feet 

91. Heating surface, tube, square feet 

92. Heating surface, total, square feet 

93. Length of grate 

94. Grate area, one boiler 

95. Ratio of heating surface to grate area 

96. Volume of steam space, water 6" above tubes 

97. Water surface, square feet 



288 ENGINEERIlSrG MECHAI^ICS. 

98. Area through tubes, square feet 

99. Area over bridge wall, square feet 

100. Volume of furnaces and combustion chambers above grates 

101. Combustion chamber head, curved or flat 

102. Combustion chamber bracing 

103 

104 

105 

Questions and Problems. 

Find the number and type of cylindrical boilers for a battleship : 
given dimensions of space available in the ship 72 feet long x 39 feet 
6 inches wide X 20 feet high; I. H. P., 11,500; pressure per gauge, 
180 pounds; air pressure allowed 1 inch of water. Find the total 
grate surface necessary and show its distribution among the several 
boilers. Find diameter and length of boilers. 

Find the number and type of cylindrical boilers; the total grate 
surface necessary, and show its distribution among the several boil- 
ers. Given: Space available 128 feet long x 35 feet wide x 20 feet 
high; I. H. P., 21,000; steam pressure, 180 pounds; air pressure 
allowed, 1 inch water. Find also the diameter and length of each 
boiler. 

From the preliminary calculations it is found that eight double- 
ended boilers will be required for the power, about 21,000 I. H. P. 
for a given ship. Length of boiler 18 feet; diameter (outside), 
15 feet 9 inches; each two adjacent furnaces have a common com- 
bustion chamber; grate surface, each boiler, 168 square feet; steam 
pressure, 180 pounds per gauge; outside diameter of tubes, 2J 
inches; length, 6 feet 6 inches. Find the number and size of fur- 
naces; thickness of shell, strength of joint being 85;^ of solid plate; 
number of tubes ; total heating surface. 

From the preliminary calculations it is found that three double- 
ended and two single-ended boilers are necessary for the power, 
about 11,500 I. H. P., for a given ship. Length of double-ended 
boilers, 21 feet; diameter (outside), 15 feet 8 inches; each two 
adjacent furnaces have a common combustion chamber; grate sur- 



BOILER PROBLEM. 289 

face, total (all boilers), 685 square feet; steam pressure, 180 pounds 
per gauge; outside diameter of tubes, 2 J inches; length of tubes, 
7 feet 7f inches. Find, for the double-ended boilers, the number 
and size of furnaces; thickness of shell, strength of joint being 85^ 
of solid plate ; the number of tubes ; the total heating surface. 

Given the formula, p= ^^^ , = 14,000, find the thickness and 

make a sketch of a Morison furnace 39 inches mean diameter for a 
boiler carrying a working pressure of 180 pounds per gauge. 

Given a boiler 16 feet outside diameter; thickness of shell plate, 
IJ inches; tubes, 2 J inches, outside diameter; pitch of tubes, 2-| 
inches; percentage strength of joint between front tube sheet and 
upper head sheet, 65fc; thickness of front tube sheet, -J inch; of 
upper head sheet IJ inches. Find the radius of the curve of the 

upper head sheet. E= ^^^^^, = 13,000. Working pressure, 

160 pounds per gauge. 

Show by neat sketches the method of securing a large double- 
ended boiler in a ship; boiler axis to be fore and aft. Show ram- 
ming chocks and show in detail the method of securing the boiler 
to the saddles. 

Design the girder stays for a flat top combustion chamber of a 
boiler; working pressure, 180 pounds per gauge; depth of com- 
bustion chamber, 30 inches ; width of chamber, 5 feet. Each girder 
to be made of a pair of steel plates 1 inch thick, carrying three stay 
bolts. The pitch of the bolts to be 7^ inches in each direction. Find 
the depth of the stays and the diameter of the bolts : f * for girders 
= 9000; it for bolts = 8000. 

The sketch represents the lower part of the front head of a boiler. 
The space around the manhole is to be supported by three stays, 
A, B and 0. Of these A is a through stay, perpendicular to the 
head ; B and are connected to the shell by palms, 5 feet back from 
the head; it for stays = 8500 pounds per square inch. Oalculate the 
diameter of each stay; pressure, 180 pounds p. g. 

Deduce expressions for the percentage strength of a riveted joint, 
^, ^1 and ^2- Given p = greatest pitch of rivets; d = diameter of 
rivets ; A = area of one rivet ; n = number of rivets in greatest pitch ; 
c = constant =1 for rivets in single shear, =1.75 for rivets in double 
shear; tensile strength of plate = 28 tons; shearing strength of rivets 
= 23 tons; t = thickness of plate. 



290 



ENGIITEERING MECHANICS. 



Given ^= 



lOO(p-d) 



and i. 



100x23xAxnXc 



find values 



p ^ 28xpxt 

of d and p, when n, c and t are known, and percentage strength of 
plate = percentage strength of rivet. 

Ordinary zigzag riveting, given p and d, find an expression for the 
distance Y between rows of rivets. 

Design a double butt-strapped joint, treble riveted, with alternate 
rivets omitted in outer row. Strength of joint to be 85^ of that of 




/^e//^A^o/e /Sx// 



solid plate; thickness of plate, 1 inch. Find diameter and pit<3h of 
rivets, distance between rivet rows and total width and thickness of 
butt straps. 

Zigzag riveting with alternate rivets omitted in outer row, deduce 
an expression for the distance V between rivet rows in terms of 
d and p. 

Design a double-strapped butt joint, treble riveted, with alternate 
rivets omitted in outer row. Strength of joint to be 80^ of that of 
solid plate ; thickness of plate, f inch. Find diameter and pitch of 
rivets, distance between rivet rows and total width and thickness of 
butt straps. Use nearest ^g- inch for detail results. Check per- 
centage of plate and rivet after finding all dimensions. 



CHAPTER XVIIT. 

Screw Propellers.* 

135. Data Given. 

Name of ship. 

Number of screws. 

Number of blades. 

Kind of screw. 

Adjustment of pitch, each side. 

Displacement (tons). 

Draft, forward. 

Draft, aft. 

Draft, mean. 

Total I. H. P. 

Eevolutions per minute. 

Speed in knots. 

Diameter of screw shaft. 

Thickness of sleeve on shaft. 

Metal of screw. 

Working face moved aft on center line. 

Blades bent back, or center line bent aft. 

Scale of drawing. 

(For calculating the diameter by comparison, use the propeller 
of the U. S. Steamer '' Olympia," d = 14'9", p = 13,500, v = 20 
knots, r= 129.) 

NOTES ON" PROPELLER DESIGN". 

The following notes are compiled from : 
Seaton : ^' A Manual of Marine Engineering.^^ 
Seaton and Rounthwaite : " Pocket-book of Marine Engineering." 
Chase : " Screw Propellers and Marine Propulsion." 
Barnaby : " Marine Propellers." 
Foley : " Mechanical Engineer's Reference Book." 
Practice of the Bureau of Steam Engineering of the Navy De- 
partment. 

In all cases, the Department method is given preference 

* From "Notes on Machine Design." 



292 ENGINEERING MECHANICS. 

DEFINITIONS. 

Chase, p. 132 : The diameter of a screw propeller is the diameter 
of a circle swept by the tip of the blades. 

The pitch is the axial distance between the same thread at one 
revolution. 

The disc area is the area of the circle described by the tips of 
the blades including the area of the hub. 

The developed or helicoidal area is the actual area of the work- 
ing face of the blades when flattened out on a plane. 

The projected area is the area of the projections of the blade 
on a plane taken at right angles to the axis of the screw. 

The length of the screw is the greatest length of the blade 
measured parallel to the axis. 

The speed of the screw is the pitch multiplied by the number 
of revolutions. 

The pitch ratio is the quotient arising from dividing the pitch 
by the diameter. 

A screw propeller is a true screw when its pitch is uniform; 
and an expanding pitch screw when the pitch increases from the 
forward to the after edge, or radially, or both. 

The principal feature of the Griffiths screw is the large boss, 
which, while not impairing the efficiency, enables the blade to be 
fixed in such a manner that the pitch can be readily altered. 

The modified Griffiths screw propeller is generally adopted for 
vessels of the navies of England and the United States, and is 
one having blades of any curved shape diminishing in width 
towards the tips and having the broadest part nearer the boss than 
the tips. 

Chase, p. 139 : " The shape of the blade is sometimes oval and 
sometimes resembles the section exposed by cutting a pear longi- 
tudinally. When made pear-shaped, it is for the purpose of get- 
ting the greatest possible surface near where the pitch angle is 45°, 
as that is considered the most efficient angle; and by the blade 
becoming diminished in width towards the tips, the surface friction 
is thereby reduced, and room is obtained for the access of the 
supply water." 

CALCULATIONS. 

Chase : " In commencing the design of a propeller for a given 
ship, the diameter and pitch are first decided upon." 



SCREW PROPELLERS. 293 

Chase, p. 130 : " The true secret of success in designing a screw 
propeller lies in the correct proportion of the three cardinal ele- 
ments — diameter, pitch and blade surface; and that the shape of 
the blades (within certain limits), their thickness, and the con- 
dition of their surfaces, are of minor importance." 

PITCH. 

Seaton, p. 329: "Let 8 = speed of ship in knots; E = number 
of revolutions per minute; s = slip in knots, generally expressed 
as a percentage, x. Then S = speed of screw, in knots per hour 



multiplied by (l--i^)- 



Speed of screw, in knots per hour: 



1-T¥¥ 100-x 

But speed of screw, in knots per hour = ^ ^tt^tt^ 

dOoO 

. pitch X revs. X 60 _ SXIOO 

6080 ~ 100-x • 

rp. ., -, 8x100x6080 S ,, 10,133 

Then pitch = ^^-p,-,^^ . = -D- X 



60E(100-x) R 100-x 

Chase, p. 134 : " The percentage of slip now allowed is from six- 
teen to twenty in large screws for good-sized ships, in which the 
engines make from 115 to 150 revolutions per minute." 

Seaton : " x is generally from 20 to 25,^ for naval vessels." 

From 24 of the latest fast vessels of the United States with 
engines of Bureau design, the average slip is 20.77^. 

Seaton, p. 323 : " If the efficiency of the mechanism were the 
same at any number of revolutions, a fine pitch propeller might be 
used to advantage, as being more efficient than a coarse one; but 
since the efficiency of the engine is much higher at a moderate than 
at a maximum rate of speed, the total efficiency of the screw and 
engine is often improved by increasing the pitch of the screw." 

PITCH RATIO. 

Chase, p. 137: "For torpedo-boats and other light-draught 
boats, which usually run in smooth water and where high-rotative 
velocities prevail, diameters being restricted in consequence of small 
draught, the pitch of screws varies from about 1.4 to two times 



294 ENGINEEEING MECHANICS. 

the diameter. For the larger high-speed ocean vessels, which 
traverse water more or less rough and where ample diameter and 
immersion can be obtained, the pitch varies from 1.2 to 1.6 times 
the diameter. For slower cargo vessels, etc., the pitch varies from 
1.1 to 1.3 times the diameter." 

Foley : " As a rule the pitch ratio should lie between 1 and 1.5 
times the diameter; 1.25 times the diameter is usually satisfactory 
in practice." 

From 24 vessels of the U. S. Navy with engines of the latest 
Bureau design, the pitch averages 1.23 times the diameter. 

EXPANDING PITCH. 

Chase, p. 32 : " The alleged advantages for a propeller of ex- 
panding pitch are that, as the screw advances through the water, 
the forward portion encounters a resistance due to a solid body 
moving through water at rest; but water, being an exceedingly 
mobile substance, that with which the screw first comes in con- 
tact has motion imparted to it by the pressure exerted by the screw ; 
and this motion of the water is accelerated by the portion of 
the blade following, and consequently the after-portions of the 
screw will press upon the water with a motion opposite to that in 
which the screw is advancing. In order, then, to equalize the 
pressure over the whole blade, the pitch must be increased pro- 
portionately to the motion acquired by the water. The increase in 
pitch usually employed is from 15 to 25^." 

It is questionable whether there is any advantage in expanding 
pitch. 

DIAMETER. 

Seaton, p. 323 : " The size of a screw depends on so many 
things that it is very difficult to lay down any rule for guidance 
and much must always be left to the designer, so as to allow for 
all the circumstances in each case." 

The diameter of the propeller must be less than the draught 
of the ship, to prevent racing and striking ground. 

Barnaby, p. 34 : " It is very important that a propeller should 
have sufficient immersion, since, if it breaks the surface of the 
water, the efficiency is reduced to a remarkable extent; but, if 
it is sufficiently far below the surface to prevent its drawing air, 



SCREW PROPELLERS. 



295 



1 <c 

TO" 



any further immersion within the limits that can practically be 
obtained is of little vahie." 

Chase, p. 140 : " The depth of immersion of the upper tips of 
the blades should increase with the draught of the vessel or the 
diameter of the ^Dropeller. It should not be less than in the fol- 
lowing table when the vessel is at the load line : 

For a diameter of 4 ft. or less tt of diameter. 

" 4 to 8 ft 

" 8 to 12 ft i 

" 12 to 16 ft i 

" 16 to 20 ft 1 

" 20 ft. or over -J 

From practice : For large vessels there should be from 1-J to 2 
feet of space from the bottom of the keel to the bottom of the 
blades, and from 4 to 4r| feet from the surface of the water to the 
blades." 

Seaton, p. 323 : " The diameter of the screw must bear relation 
to the turning power of the engine; for, if made of too small 
dimensions, it cannot absorb the power of the engine, however 
coarse the pitch may be." 

Since the area of surface on which the skin resistance acts in- 
creases as the square of the distance from the center, and the arm 
on which it revolves increases with the distance from the center, 
the frictional resistance per square foot of surface varies as the cube 
of the distance from the center, so that any increase in diameter 
means a large increase in resistance. 

If a screw is of sufficient diameter and has ample blade area, 
any addition beyond the tip will act then as a brake and seriously 
reduce its efficiency. 



Seaton, p. 324: "Diameter of screw = K J j^ 



H. P. 



(PxR)^ 
" I. H. P. = total— 6 to 10^ for auxiliaries. P = pitch in feet. 

Rr= revolutions per minute. K = constant." Seaton gives for K a 

value of 20,000. 

From trials of similar ships the value of K is determined for 

the one being designed. 

Seaton and Rounthwaite- give a value of 22,700 for naval vessels 

of fine lines. 



296 ENGINEERING MECHANICS. 

The following are the average values of K : 

For large battleships, K= 19,000. 

For large armored cruisers, K= 22,768. 

For small cruisers and gunboats, K= 22,000 (about). 

For torpedo-boats and destroyers, K= 27,000 (about). 

PERIPHERAL SPEED. 

Foley : " Owing to the great loss of efficiency from friction at 
high speeds, it is usual not to exceed about 6500 feet per minute 
at periphery.'^ 

This should be applied as a check after the diameter is calculated. 

DIAMETER BY COMPARISON. 

Barnaby, p. 69 : "To find the diameter of a propeller for a 
given I. H. P. and a given speed from the diameter of another 
similar propeller at a different I. H. P. and at different speed : 

If d= diameter of model, which may be larger or smaller 
than D ; 
D=: diameter of required propeller; 
p = L H. P. model; 
P = I. H. P. of required propeller; 
v= speed of vessel with model propeller; 
y = speed of vessel with required propeller ; 
r = revolutions per minute of model propeller; 
E = revolutions per minute of required propeller. 

ThenD = y^d2x-^X-^ and R = rX -^ + -^ . 

The pitch ratio must be the same as that of the screw, which 
is treated as a model." 

AREA 

Seaton, p. 329 : " The best area is not easily determined, except 
by experiment by data derived from the performance of similar 
ships with similar screws. 

Good results are obtained by : 



Total area of screw blades = K 



/I. H. P. 



SCREW PROPELLEES. 297 

Single screws. Twin screws. 

The value of K for four bladed is. . 15 10.5 

three " .. 13 9.0 

'' '' two " .. 10 

Chase, p. 158: "For naval vessels with twin screws K=7 to 
8.5.^' 

Seaton and Eounthwaite : " For very fine-lined naval vessels with 
three-bladed screws K = 7.75.'' 

From 21 vessels of the U. S. Navy with engines of latest Bureau 
designs K= 8.847. 

For large battleships K==11.0. 

For large armored cruisers K=:9.0. 

Seaton, p. 330: "The projected area is often taken as the 
criterion, and is to some extent a true one.^^ 

AREA FROM DISC AREA. 

Foley : " The developed surface is usually from 28 to 40^^ of the 
disc area. For ordinary cases of moderate or high speeds, where the 
diameter is well proportioned, 35^ is usually satisfactory. It should 
be remembered that there may be great loss of power from too much 
surface." 

For the disc area the boss is usually spherical, and from J the 
diameter of the screw for small screws to J the diameter for large 
ones. 

AREA EROM THRUST. 

217 

Seaton, p. 201 : " P = I. H. P. x ^ . P is mean normal thrust; 

K is speed in knots per hour." 

" P varies with the I. H. P. and inversely as the speed in knots, 
so the thrust may vary considerably. If the speed is from any 
cause reduced the thrust must increase." 

"To allow for the variations of P, the surface should be such 
that the pressure per square inch from the mean normal thrust does 
not exceed 70 pounds." 



Indicated thrust= ^- ^- P.X13.000 



revs, per min. x pitch in feet * 

"The efficiency of the engine and propeller are each taken at 
75^. 



298 ENGINEERING MECHANICS. 

Then, LH. P. x 33,000 becomes I. H. P. x 33,000 x. 75 x. 75 
= LH. P. X 18,562/^ 

From six of the vessels of the U. S. Navy with engines of the 
latest Bureau design, the average thrust per square foot of helicoidal 
area is 1410.17 pounds, which is much later practice than that of 
Seaton. This is for propellers of manganese bronze. 

The area from thrust is to be a check on the other methods. 

BLADES. 

Chase, p. 135 : ^^ Steadiness in action determines the number of 
blades. For a single screw, four blades give the greatest satis- 
faction, especially in rough water.^' 

" Two-bladed propellers cause too great vibration.^' 

" With twin screws where the supply water is not so much, inter- 
fered with, and where the immersion is more ample, three-bladed 
screws are preferable.'^ 

Seaton and Eounthwaite : " Three blades are generally used 
when the immersion is sufficient and the revolutions fairly high; 
but when the calculated diameter of the propeller becomes nearly 
as great as the draught of water, and the revolutions are only 
moderate, four blades should be used.'' 

Seaton : " If the necessary disc area can be obtained with a 
three-bladed propeller, this form is preferable, as less shock is given 
to the ship; but if the draught is limited, a four-bladed propeller 
may be necessary." 

Practice: Blades are generally separate from the hub in naval 
vessels. 

MATERIAL OF BLADES. 

Seaton, p. 335 : '^ Cast iron is generally used because cheap, and 
when struck breaks clean off and goes out of the way. 

" Cast-steel blades are largely used, being thinner and lighter for 
the same strength than cast iron. 

" Wrought-steel blades are strong and efficient. The chief objec- 
tion to steel is that it corrodes rapidly on the back of the blade, 
due probably to air detached from the water. 

^' Phosphor and manganese bronzes are largely employed for 
making propellers for naval vessels. The blades can be cast thin- 
ner than if made of steel, as there is little corrosion. They are 
objected to on the ground that they cause galvanic action, and 
injure the iron or steel near by; but this is questioned." 



SCREW PROPELLERS. 299 

Seaton and Eounthwaite : " For naval vessels propellers are al- 
most invariably of gun-metal or bronze^ and except sometimes in 
torpedo-boats, the blades are always detachable." 

Bureau Practice : " For U. S. naval vessels manganese bronze is 
always used in the latest designs." 

FORM OF BLADES. 

Seaton, p. 333 : " The forms of blades are numerous and varied. 
Some bend the blades forward, some back; but no general plan 
has been found to give, always, the best results. The better results 
are due to a better proportion of blade rather than to the shape." 

Barnaby, p. 78 : " The standard blade area of the Admiralty 
standard blade is an ellipse of major axis equal to the radius of 
the propeller and of minor axis equal to four-tenths of the major 
axis. It is often found, however, that owing to the diameter being 
limited, sufficient blade area is not obtained by these proportions. 
In such cases the elliptical form is adhered to with an increased 
minor axis of from .5 to .55 of the major." 

LENGTH OF BLADES. 

Chase, p. 140 : " This is important and is determined by practice. 

"The length affects the power to move a given columnal mass 
of water. When made too long, the loss from the divergent and 
rotary direction given the discharge water and also from the surface 
friction of the blades more than counterbalances the advantage of 
the increased mass of water acted upon. 

" The length should vary with the rotative speeds. The most 
efficient length seems to lie between 18 and 20fo of the diameter — 
the former for high and the latter for low speeds. The average 
of a number of high-speed screws of modified Griffiths shape gives 
a maximum length of 19^." 

BREADTH OF BLADES. 

Seaton, p. 334: " The greatest breadth of blade should not, as a 
rule, be beyond one-third the radius of the disc from the center, 
and should be approximately as given by the following rule : 



3/j JJ p 

Max. breadth of blade in inches = K \ ' — '- . 

▼ Rev. 

"For a four-bladed screw, K = 14; for three-bladed, K=17; and 
for two-bladed, K=:22. 



300 ENGINEERING MECHANICS. 

" The breadth of blade at the tip should be from one-third to 

two-fifths of the maximum/' 

THICKNESS OF THE BLADE. 

To find the thickness of the blade at the center of the shaft (that 
is, supposing that the taper of the blade section were continued to 
the shaft axis) : 

Let b = maximum breadth of blade, inches. 

n= number of blades. ♦ 

R= number of revolutions per minute. 
f = 90. 
K=2.56. 



/I H P yf 

Then, thickness of blade (at center of shaft) =a/ -^^^ — '-^ xK. 
^ ^ y Exnxb ' 

where f may be taken as 100 for compound engines of the ordinary- 
two and four-cylinder type; 90 for three and six-cylinder com- 
pound engines, and 85 for three-crank triple-compound engines. 

"The value of K is 4 for cast iron; 2 for ordinary gun-metal; 
1.6 for cast steel and 1.5 for forged steel and bronzes of superior 
make.'' 

Bureau Practice: "From 24 vessels of the U. S. Navy with 
engines of the latest Bureau design, the average value of K is 2.56." 

Chase, p. 162 : 
" Cast iron allow .5 in. for each foot in dia. of screw. 

Cast steel " .4 

Gun metal " .45 

High class bronze. .. " .4 

THICKNESS AT THE TIP. 

Seaton, p. 330 : " The thickness of metal at the tip should be 
.2 of that at the root." 

Chase, p. 160 : " D is the diameter of the propeller in feet. 

Thickness at the tip, cast iron 04D + .4 in. 

" " cast steel 03D-I-.4 in. 

« " gun metal 03D + .2 in. 

" " highest class bronze 02D + .3 in. 

Also, Chase: "The thickness at the tip should be as small as 
possible consistent with good casting or forging. When gun-metal 
is used it is generally made about -^\ inch for each foot in diameter 
of the screw." 



SCREW PROPELLERS. 301 



BOSS OR HUB. 



Seaton, p. 332 : " When a loose-bladed propeller is designed, 
the boss is usually spherical in general form with flats or recesses 
for the blades. The diameter of the sphere is from J the diameter 
of the screw for small propellers to -J the diameter for large ones." 

Chase: " Boss = diameter divided by 3 J or 3 J, which will cut 
away about one-fifth of the standard ellipse." 

Seaton, p. 332 : " The length of the boss depends on the size of 
the base of the blades, and is generally about .85 of the diameter 
of the sphere for two-bladed screws and .75 for four-bladed. 

" In most navies gun-metal is used for the hubs of propellers." 

BOSS AND FLANGES ON BLADES. 

Chase : 

"Diameter of flange = 2.25 x diameter of shaft. 

" Thickness of flange = .85x diameter of stud for bronze or steel. 

" Length of boss = 2.7 x diameter of shaft. 

"Thickness of boss = .65 x thickness of blade at shaft axis. 

"Diameter of boss = 3.3 X diameter of shaft. 

" Taper of boss = l" to 1'. 

" The fore and aft section of the boss should be oval — the prin- 
cipal radius being .8 x diameter of boss." 

Seaton : " A well-designed and carefully made screw should 
have the base of the blade conforming to the general outline of 
the boss, and the nuts or bolt heads recessed into the blade base 
and covered with a metal case or cemented flush with the surface." 

BOLTS. 

Chase, p. 161 : " The bolts are usually of naval brass (muntz- 
metal with the addition of a small percentage of tin), or are of 
one of the stronger bronzes for gun-metal or bronze propellers, and 
of mild steel for cast-iron or cast-steel propellers. 

TxL 

" The size is from the formula : a X N" x r = — -^ — . 

a = area in square inches of one stud bolt at bottom of thread. 

N=:Xo. of studs or bolts for one blade (usually 7 to 11). 

r = radius of pitch circle of studs or bolts in inches. 

T = indicated thrust per blade 

^ I. H. P. X 33,000 

~" mean pitch x revolutions per min. x No. of blades * 
20 



302 E^TGINEERTNG MECHANICS. 

L = .6x total length of blade in inches (flange joint to tip). 
X = 1700 for steel bolts or stnds. 

1400 for naval brass or bronze studs or bolts. 

TAPER OF SHAFT, FEATHER^ THREAD^ NUT. 

Chase, p. 161 : " The taper of the part of the shaft within the 
boss should be about one inch in diameter for each foot of length, 
but never less than three-fourths of an inch. 

"The thread of the large nut that holds the propeller on the 
shaft is 2^ threads per inch, regardless of the diameter. It should 
be left handed when the propeller is right handed, and vice versa. 

" The nut should be securely locked, preferably by a plate fixed 
to the after end of the boss by set screws. 

" The propeller should be secured by one feather or key extend- 
ing the whole length of the boss, the proportions of which may be : 
Breadth of key = .22 x largest dia. of shaft + "25. 
Thickness of key == .55 x breadth. 

" The diameter of the screwed end of the shaft should be suffi- 
ciently reduced to allow the key to be fitted in from the after end 
clear of the thread.^^ 

The key should have clearance at the after end, so that the nut 
will not come tight against it, and should be partly imbedded in 
the shaft, the forward end being rounded. 

In Bureau draiuings, sometimes the key is arranged to be secured 
in place in the shaft and the propeller slid on from aft, in which 
case the top of the thread on the shaft should be less in diameter 
than the smallest bore of the propeller. 

ALTERING THE PITCH. 

Barnaby, p. 91 : " For the sake of pos- 
sible adjustments that may be desired on the 
trials of machinery, and also for finding 
the pitch when it becomes necessary to 
reduce the steam pressure in the boilers, it 
is customary to make the bolt holes in the 
blade flanges oval, in order that the incli- 
D E. B c, nation of the blades to the axis may be 

altered. The amount of the oval may be determined thus : ^ Let 
the true pitch of the blade be 27rCE and let the desired range 




SCREW PROPELLERS. 303 

of pitch be SttEB and 27rED respectively on each, side of this pitch. 
Take a radius CA, so that A is about half way up the blade. Join 
AD, AE, AB. Then, if c/) = the angle DAB, the holes in the flange 
must be so elongated as to admit of the blade being turned through 

-~ deg] ees on each side. The amount of this elongation on each 

side is, then, - 'j^ _~ x — ?- = ttC ^qTt ' ^ being measured in degrees. 

2 X this length + the diameter of the bolt or stud = length of slot. 
C = diameter of bolt circle.^ " 
136. Calculations Required. 
Percentage of slip. 
Pitch, mean. 
Pitch, greatest. 
Pitch, least. 
Pitch, ratio. 

Diameter from pitch ratio. 
Depth of immersion (Chase). 
Depth of immersion (Practice). 
Distance below blades. 
I. H. P. of auxiliaries. 
I. H. P. of each engine used in calculations. 
Value of K USED^. 
Diameter, Seaton. 
Diameter, Foley. 
Diameter, Foley, 2d method. 
Diameter by comparison. 
Average of methods. 
Peripheral speed. 
DIAMETEE USED. 
Value of K used. 
Helicoidal area, Seaton. 
Diameter of boss, Foley. 
Diameter of boss. Chase. 
Diameter of boss, Chase, 2d method. 
DIAMETEE OF BOSS, USED. 
Disc area. 

Helicoidal area from disc area. 

Thrust, Seaton. i 

Thrust, Foley. 



304: EXGIXEEKING MECHAXICS. 

THEUST USED. 

Helicoidal area from thrust, Seaton. 

Helicoidal area from thrust, Bureau. 

Helicoidal area, average. 

HELICOIDAL AEEA USED. 

Blades, number. 

Blades, material. 

Bend back of blades. 

Helicoidal area of each blade. 

Major axis of trial ellipse. 

Minor axis of trial ellipse. 

Major axis of ellipse USED. 

Minor axis of ellipse USED. 

Length of blade, Chase. 

Length of blade, Seaton. 

lejstgth, used. 

Distance of maximum breadth of blade from center. 

Breadth of blade at the tip. 

Thickness of blade at IJd, Seaton, 1st method. 

Same, Seaton, 2d method. 

Same, Chase. 

THICKNESS OF BLADES, USED. 

Thickness at tip, Seaton. 

Same, Chase, 1st method. 

Same, Chase, 2d method. 

THICKNESS AT TIP, USED. 

Boss, diameter. 

Boss, length, Seaton. 

Boss, length, Chase. 

BOSS, LEXOTH USED. 

Boss, thickness. Chase. 

Boss, taper. 

Boss, radius. 

Flange, diameter. Chase. 

Flange, thickness. Chase. 

Bolts, material. Chase. 

Bolts, number, Chase. 

Bolts, diameter. Chase. 

Bolts, diameter, Chase, 2d method. 

Taper of shaft. 



SCREW PROPELLERS. 305 

Threads per incli on shaft. 

Eight or left handed. 

Breadth of key. 

Thickness of key. 

Length of key. 

Diameter of screwed end of shaft (outside). 

Depth of thread. 

No. of threads to 1". 

Diameter of plug in the end of shaft. 

Length of same. 

No. of threads to 1" of same. 

Length of slots in flanges of blade. 

137. Geometry of the Screw. — It is first noted that the face only 
of the blade is drawn of the true geometrical form and is the driv- 
ing face of the blade; the form of the back depends on the thickness 
given at the different radii. 

An exception is sometimes made in the case of torpedo-boats, 
where rapid manoeuvring is required and rapid speed astern, where 
the front and back of the blades are alike. 

Barnaby, p. 175 : " If a point moves on the surface of a cylinder 
in such a way that, while moving uniformly around the cylinder, 
it advances uniformly in the direction of its axis, it will trace 
a curve known as the helix. Imagine the cylinder cut on one side 
by a straight line, BC, parallel to the axis meeting the helix in 
consecutive points A and C, and then un- 
rolled and laid flat, the circumference through A 
A will become a straight line AB; BC at // 
right angles to it will represent the direction / / 
of the axis, while that part of the helix // 
formed during the complete revolution of the / / 

tracing point will be represented by the line —^ — ^- 

AC. Since the distances moved by the point 
in the directions AB, BC are proportional, AC is a straight line; 
BC, being the distance moved through in the direction of the axis 
while the point goes once entirely round the cylinder, is the pitch; 
while the angle BAC, which the unrolled helix makes with the plane 
at right angles to the axis, is the angle of the helix or screw. 

" If a straight line moves uniformly around an axis which it 
intersects, and to which it is always at right angles, advancing at 
the same time uniformly in the direction of the axis, it will sweep 



306 EXGIXEEEIXG 3IECHAXICS. 

out a surface known as a helicoid, and every point in the gen- 
erating line will describe a helix as shown above, necessarily lying 
on this helicoid. Since, during a complete revolntion of the generat- 
ing line^ every point moves throngh the same distance in the direc- 
tion of the axis, the helicoid is the surface of uniform pitch; that 
is, BC is constant for the helices of all points in the generating line. 
A helicoid can, therefore, be, and often is, used for the acting face 
of a screw blade of uniform pitch. It is not necessary, however, 
that the generating line should be at right angles to the axis; such 
a surface ma}^ be generated by any line, straight or curved, moving 
uniformly along and revolving uniformly around an axis, intersect- 
ing and always making the same angle with it. 

'' The helix traced by any point in the generating line will also 
be the curve of intersection with the screw surface of a coaxial 
cylinder of radius equal to the perpendicular distance of the point 
from the axis. The larger the radius of the cylinder, the larger of 
course the length of the circumference, as A'B ; and, as the pitch is 
constant, it follows that the angle of the helix must decrease as the 
radius of the intersecting cylinder increases. We thus arrive at the 
fundamental geometrical property of a surface of uniform pitch, 
viz : Co-axial cylinders intersect it in helices, all of which have the 
same pitch, but whose angles vary, decreasing as the radius of the 
cylinder increases. Near the axis, therefore, the helices will ap- 
proximate in direction to that of the axis, and as the distance from 
the axis increases, they will lie more and more at right angles to it. 
If 6 be the angle of the helix, p the pitch and r the radius of the 

intersectinsT cvlinder, tan ^= ■—— .^^ 
° " ' 2-r 

Barnaby, p. 78 : '^ The expanded blade area, which may be 
described as a flat surface of approximately equivalent area to that 
of the blades, both as to amount and disposition, is derived as 
follows : 

"A co-axial cylinder will intersect the screw surface in a heli- 
cal curve making a certain angle with the axis, and it will inter- 
sect a plane passing through that diameter of the cylinder which 
passes through the middle point of the helical curve and making 
the same angle with the axis, in an elliptical arc. The length of the 
screw being small compared with the pitch, these two arcs will 
nearly coincide, and no great error will be involved by assuming that 
they do coincide. Imagine these elliptical arcs at all radii to be 



SCREW PROPELLERS. 307 

swung around a common center line till they all lie in the same plane 
with their major and minor axes respectively coincident (though 
necessarily of different lengths), then the curve passing through 
their extremities will form the expanded blade area. This area is 
very nearly equal to the actual working face of the blade, being in 
fact somewhat less than it." 

Drawing. — The following is a general form for laying down the 
screw to be observed in the designing-room of the Department of 
Marine Engineering and Naval Construction at the United States 
Naval Academy: 

The drawing is to consist of vertical elevation of the propeller 
from aft; one-third of the hub and one flange of the blade in 
section; a side elevation; a side elevation of one blade showing 
the section of the hub and the thickness of the blade; a draw- 
ing of the developed area and sections of the blade either on the 
angular lines or at right angles to the central line of the blade 
on one of the other figures; guide-irons and sections of flanges and 
bolts as required. 

The side elevation of the blade showing the sections of the hub 
may be combined with the side elevation of the whole propeller 
by making part section and part elevation, though it is better to 
separate them for clearness. 

Having determined most of the dimensions from the previous 
notes, we first begin Fig. 1, Sheet 1, a vertical elevation of the 
blade looking forwards. Draw the center line OPO' represent- 
ing a line at right angles to the center line of the shaft, 0. Lay 
off the circle showing the greatest diameter of the hub, and draw 
a vertical center line through 0, the center of the shaft, for the 
center line of the blade. Lay off OA equal to the radius of the 
propeller. Lay off equally on each side of this line the developed 
area of one blade required. Since about one-fifth of this area 
will be inclosed in the hub, it is well to lay off at first the area 
on OA as an axis equal to six-fifths of the area required. This 
area may be laid off entirely by trial suiting the curves to the 
ideas of the designer, but it is better to use as a basis for a trial 
area an ellipse with OA as a major axis and a minor axis found 
as follows: 

Area of an ellipse = vrab, when a and b are the semi major and 
minor axes. In this case, the area must equal f X developed area 
calculated for one blade. The major axis is equal to OA. To find, 



308 



ENGr^"EEllI^^G mechanics. 



then, the minor axis, we have Trab =f x helicoidal area of one blade. 
Then, 

6 X helicoidal area of one blade 



Svra 



=b. 



! ^/-/-7^— ^->-^,A. 




Fig. I. 



Fici.2. 



Sheet 1. 



The ellipse is then constructed according to any approximate 
method, and the area TE^ AE^^ UHT is measured by planimeter. 
If the area does not agree with that required, it may be enlarged 
or diminished by changing the length of the minor axis until the 
required area is obtained. If the area be too small, one way of add- 
ing to the surface is to enlarge the lower part, thus making the blade 



SCREW PROPELLERS. 309 

more pear-shaped, according to the ideas of the designer. If the 
area be too large, sometimes a portion is removed from the top of 
the blade, thus again producing a pear-shaped blade. 

In Fig. 1, Sheet 1, the lower part of the blade has been made pear- 
shaped, the lower portion of the original ellipse not being drawn. 

Some designers use for the semi-major axis of the ellipse about 
S5^i of the radius of the propeller, and take four-tenths or more, as 
required, for the semi-minor axis. The ellipse then is not so much 
within the circle of the hub and a less proportion is cut off. 

Divide the center line HA into any number of parts — five or 
six, and preferably equal, or, as generally constructed, make the 
outer divisions, AB, BC, etc., to FG equal and a definite number 
of feet and half feet long and the divisions GH nearest to the 
hub longer or shorter than the others, if necessary. Draw horizontal 
lines through these points. 

Lay off 0P=£^ and draw lines through PA, PB, etc., to 

PH. Tane=^-^f^~=^ = ^. 
Circumference 2kT UA 

r is a varying quantity, OE, OD, etc., to 
OA. Since the tangent of the angle which 
any one of these lines PA, PB, etc., passing 
through a point in OA distant r from ^ ^ Ortumfervnce 

makes with AO is ~- , these lines represent the angles of the 

blade of the various sections at distances OB, OC, etc., from the 
axis. 

From the points B, C, D, etc., set off on the lines PB, PC, etc., 
the lengths of the sections made by the horizontal lines through 
B, C, D, etc. on the expanded blades, as shown by the arcs. 

Next, in Fig. 2, Sheet 1, the side elevation, or fore and aft 
projection, draw the vertical center line O'A' for the center line 
of the hub. The blade is seldom laid off on this line, as it is 
frequently found that, if the thickness of the blade be laid off 
from this line, the back of the blade will interfere with the bolts 
used for securing the blade to the hub. Consequently, the work- 
ing face is moved back, as the thickness is placed on the forward 
side. This distance of moving back the center line of the blade 
varies with the diameter of the propeller, and in this case is IJ 




310 ENGHSTEERING MECHANICS. 



SCREW PROPELLERS. 311 

inches. The lines O-Aj, 1| inches back^ is then the center line of 
the blade. 

From previous calculations we have the thickness of the blade 
at the tip and at the center of the shaft. Lay off from O^A^ these 
distances (the one at the hub, t, and the one at 1^ diameter of 
shaft, i', will be checks on each other), and connect the points of 
thickness at tip and hnb by a straight line, the extension of this 
line (dotted) to the thickness at l-|d, being drawn. 

Where the line showing the center line of the blade and the one 
showing the thickness meet the hub easy curves are drawn as shown. 

We have then a section of the blade through the center line of 
the developed blade. 

This thin sectional strip showing the thickness at the different 
distances from the center is supposed to represent a section through 
the center line of the flattened or developed blade, as it is apparent 
that only on this assumption could a plane section be obtained 
giving the thickness of metal normal to the blade. 

Extend the horizontal lines through A, B, etc., of Fig. 1, Sheet 
1, to Fig. 2, which gives the thickness of the blade for the various 
sections represented by the horizontal lines. Transfer these thick- 
nesses to Fig. 1 by laying them off at B, C, etc., perpendicular to 
the lines representing the angles of the blades, as shown at E and F, 
and draw through these points arcs of circles touching the edges 
of the sections already obtained. These arcs should really be 
portions of ellipses, but are so near portions of circles that they 
are practically correct. This will give the shapes of the sections 
of the blade at different radii. 

In some drawings these sections are laid out horizontally on the 
lines B, C, etc. Either way may be followed. 

Those sections near the base are sometimes modified in order 
to allow the water to escape freely from the following edge of the 
blade, in which case the outlines of these sections are not finished 
until the amount of this modification is determined. The follow- 
ing edge is frequently curved upwards or back, and the amount 
of this flare is governed by the designer. The portion to be 
rounded back is shown by the assumed dotted line 12^?,-^ on the 
developed blade. By measuring the distances 22^, SS^, 44^ and 
setting them off as 2323, 8283, 4343, on the lines showing the angles 



312 ENGINEERING MECHANICS. 



SCREW PEOPELLERS. 313 

at the different distances from the center, we find the points where 
the blades begin to flare. Draw easy curves SgS^, 333^, 434^, thus 
dropping back the working edge near the hub and still preserving 
about the same helicoidal area. The perpendicular distances from 
3^, 34 and 44 to the corresponding lines for the angles of the blades 
give the actual amounts of the setting back of the edges to be used 
in the plan view. 

These points 2^, 3^ and 4^ are connected b}^ arcs of circles with 
the points showing the thickness of the blade at the center and 
the other edge as found. 

The plan, Fig. 3, Sheet 1, is next begun. A circle of the diameter 
calculated for the flange of the blade is drawn. If the working 
face of the blade were not moved back from the center line of the 
hub, Fig. 2, the projection of the central line of the blade would be 
at Oi, Fig. 3. In this case, the center having been moved back, the 
center of the blade in the plan view will be at A2;, a distance aft equal 
to that in Fig. 2. Through the point Ag draw lines representing the 
angles of the blade for the sections A, B, C, etc., as in Fig. 1, and 
lay off on these lines the developed lengths at the different sections. 
For instance, E^'E^-^ — 'E^A^E,^- N'ote that the lengths of the sections 
of the lower portions of the blade where cut away are measured to 
the actual length on the after edge of the blade, as A2Gr2 = Gr32. Then 
lay off perpendicular to AgGg the distance from the section, Fig. 1, 
that the blade is bent back from the section, as G2S=^2^4- 

Before completing the plan view, the intersection of the blade 
with the spherical surface of the flange is obtained. In Fig. 1 
it is seen that the lower edges of the developed blade, T and U 
are just within the circumference of the flange, so that the edges 
of the intersection of the flange and hub should be on this line. 
To get the intersection it is convenient to draw Fig. 4, though 
the same operation may be carried on in Fig. 3. Draw the circle 
of the flange. Fig. 4, and place in its proper position the section 
of the blade at H of Fig. 1. This section is, of course, tangent 
to the round up of the flange. It is seen that this section over- 
laps the circle of the flange, so that from this point downwards 
to the blade the width must be gradually shortened, so that the 
edges may fall within the circumference of the flange; and these 
edges must be distant from the center of the flange a distance 
equal to the radius at TU, Fig. 1. Another consideration that 



314 E]SrGINEEKIN"G MECHANICS. 



SCREW PROPELLEES. 315 

affects the positions of the points of intersection of the edges of the 
jSange and the blade is the position of the bolts, covering plates, etc. 
These are drawn in, according to experience. The number of bolts 
has already been decided upon and also their diameters, so that the 
distance apart and the clearance must be determined to the best ad- 
vantage. The outlines of the covering plates being determined, the sec- 
tion at H is worked down in Fig. 4, so as to shorten and possibly 
twist a little one way or the other as required. Drawings from the 
Bureau of Steam Engineering are to be consulted for this. The 
points M and N will be determined from the required distance from 
the center, and the best arrangement of the line of intersection of 
the flange and blade as shown by the unhatched lines, taking care 
that the greatest thickness at the middle point of the intersection is 
not less than the corresponding thiclaiess of the section H. 

This theoretical intersection is not shown on the final drawing, 
as the fillets at the bottom of the blade cause easy curves of 
intersection. 

To find where the ends of the blades intersect the flange in the 
different projections : 

The points M and N are transferred to Fig. 3, thus giving M, 
and ^2? ^^^ a^6 projected to Fig. 1 to the line TU, thus giving 
Mg and Ng. Project M and 'N, Fig. 4, to a vertical line and 
measure the distances from the center of the blade and transfer 
these distances on either side of A^Cs, Fig. 2, on the line TU 
extended for the points M^ and 'N^. 

To obtain the projections of the working edge of the blade : 

Having determined Mg and ^"2, Fig. 3, we have now all the 
points of the plan, so we may draw in the curve ISr2E2A2E2M2. 
In Fig. 1, project from Pig. 3 the ends of the lines EgAgEg, etc., 
to their corresponding lines on Fig. 1 and draw a curve through 
these points and extend to Mg and jSTg, taking care to note which 
edge falls behind the round up of the flange. 

For Fig. 2 project the ends of the angular lines of Fig. 3 to a 
vertical plane and measure on Fig. 2 from the center line A^O^ 
and on the corresponding lines the distances on this vertical plane 
from the projection of the center Ag. Connect the points so found 
with a curve, extending it to M4 and IST^, noting which curve falls 
behind the round up of the flange. 



316 



ENGINEERIlsTG MECHAlsTICS. 



Another method of finding the projections is as follows : 
In Fig. 1 project E^ verticall}^ to Eg. Measure EgE^ from E 
to Eg. This will be then one of the points of the intersection. In 
Fig. 1 measure EEg and lay it off on Fig. 2 on the correspond- 
ing line and edge of the blade from the center line O^A-^ of the 
blade. 

The same methods are followed for all points of the projections. 



OUTLINE OF THE BACK OF THE BLADE. 

In the plan, Fig. 3, Sheet 1, it 
is evident that a portion of the 
back of the blade is seen. The 
outline is obtained as follows: 
On the elements B2B2, C2C2, 
etc., lay off the thickness strips 
as determined in Fig. 1. Then 
the highest portions of the differ- 
ent thicknesses will be points in 
the outline of the view of the 
back of the blade, so that a curve 
drawn tangent to these curves 
will give the outline of the back 
of the blade. 

Sheet 2 shows this method of 
finding the outline of the back 
of the blade. At some point 
this outline gradually joins the curve of the flange. This is repre- 
sented approximately by a curve. 

Having determined the general outline of the blades for the 
different views, the hub may be drawn and finished according to 
the dimensions and details as determined upon. Some of the 
methods as used in the designs of machinery for vessels of the 
United States Navy are shown in Sheets 3 and 4. 

In the case of a screw with expanding pitch the above method 
of drawing the propeller is somewhat changed. See Sheet 5. 

entering pitch 

27r 




Sheet 2. 



Divide the center line as before and set off OP: 



and 0F= leavmg£itch^ -p^^^ p ^^^^ ^.^^^ through B, C, 

Ztt 



SCREW PROPELLERS. 



317 




Sheet 3. 



Sheet 4. 



21 



318 



ENGINEERING MECHANICS. 




Pitch of BladeJ<f€^^~- -/y^'- 



■166 
Sheet 5. 



etc., these lines giving the an- 
gles of the blades for the enter- 
ing edges. From P' draw lines 
through B, C, etc., for the 
angles of the blade for the leav- 
ing edge. These lines intersect 
at the center, and for the enter- 
ing half of the blade the lines 
from P form the bases of the 
sections cnt out, while lines 
from P' form the bases of the 
sections cut from the after half 
of the blade as shown. 

The rest of the drawing con- 
forms to this method. 

GENEEATEIX INCLINED. 

Barnaby, p. 85 : " Blades are 
sometimes made with the gen- 
erating line inclined to the axis; 
or, in technical terms, they are 
made with a skew. Let the two straight lines, AB, AC, of the 
figure, the former at right angles to the axis, the 
latter inclined to AB at an angle a, moving to- 
gether, generate screw surfaces of uniform and 
equal pitch. Then the helices of intersection of 
these two surfaces will be exactly similar, and one 
will be always a constant distance from the other ; 
this distance being at a radius r, equal to r tan a. 
Imagine these two surfaces so far similar that when 
AC at any radius leaves the surface, AB at the 
same radius leaves its surface, then the expanded area of the sur- 
face so formed by AB will represent what may be termed the 
effective expanded area of that formed by AC, and this should 
be of the elliptical or other form which would have been used if 
the generating line had been at right angles to the axis. A blade 
generated by AC would therefore be formed from a blade gener- 
ated by AB, simply by setting the helices of intersection definite 
distances aft; the distance at M, for example, being MN. It 
follows, therefore, that the athwartships projection of the blade 




SCREW PROPELLERS. 



319 



for the same " effective " expanded area is independent of the 
skew, and consequently for a skew blade with effective expanded 
area as for the blades shown in Figs. 1, 2 and 3, Sheet 6, the 
athwartships projection will be as in Fig. 1, no matter what the 
skew may be : 

" The fore and aft projection of the upper blade, Fig. 4, will 
be formed by using a center line inclined to the vertical at an 
angle equal to the inclination to the vertical of the generating 
line, and proceeding as in Fig. 2, Sheet 1, setting off the distances 




Sheet 6. 

horizontally. The projections of the lower blades are determined 
in a similar way to that described before. For example, the point 
whose athwartships projection is Cg, Fig. 1, Sheet 6, will appear on 
Fig. 4 as C4, lying in the same vertical line as c^ (C7 being the 
position of this point when the blade is upright) , and is perpendicu- 
larly away from the axis, a distance equal to that of Co from the 
horizontal plane through the axis. 

"For the horizontal projections, Fig. 5. — For the top blade the 
pitch lines are not all dra-^-n through the same point as in Fig. 3, 
but each line is drawn at a corresponding angle to the axis through 



320 ENGIIsTEERI^G MECHANICS. 

a point on the axis at a distance from M^I^i, equal to the distance 
of the corresponding point on the generating line PQ from MN, 
Fig. 4, the process then being as previously shown. For the lower 
blade we proceed as follows: The point (C2, Fig. 1) correspond- 
ing to c^, Fig. 4, when the blade is upright, will appear in Fig. 5 
as Cg at a distance from the axis eqnal to the distance of c^, Fig. 1, 
from the vertical plane through the axis and from M^^i equal to. 
the distance of c^ from MN, Fig. 4. 

"The projections of a three-bladed propeller with skew blades 
are shown in Figs. 1, 4 and 5, except that the left-hand lower 
blade of Fig. 1 has not been shown in Fig. 4 to avoid confusion. 

" Where the generating line is curved, the method is now obvious." 

138. The above described method of designing from an assumed 
elliptical form of developed area is subject to the following criti- 
cism : 

If the pitch ratio varies the form of the projection of the blade on 
the disk (that is, on the transverse plane of the ship) for a given 
developed area changes. If the pitch ratio is constant and the 
width of the developed blade is changed from the standard on which 
the experimental formula are based, in order to obtain the required 
area, the distance of the center of pressure from the shaft axis is 
changed. When the blade is wider this distance increases, when it is 
narrower the distance decreases, and the resistance of the screw to 
turning is thus increased with greater width, not only on account 
of the increased area, but also on account of the greater arm at 
which the resistance acts. When the blade is narrowed the resist- 
ance is decreased by the decrease in area and also by the shortening 
of the resistance arm. Thus the rough assumption that the thrust 
will vary inversely as the developed area for the absorption of equal 
powers is very considerably in error. 

To overcome this objection the Bureau of Steam Engineering has 
very recently adopted a standard form of projected area, which has 
been obtained by a combination of the best of a very great number of 
forms that have been carefully investigated. In connection with 
this standard form of projected blade the Bureau uses a chart of 
curves which has been developed by Commander C. W. Dyson, 
U. S. N". For a full description of the method see an article by the 
above-mentioned officer in the Journal of the American Society of 
Naval Engineers, Vol. XXII, Xo. 3, August, 1910. 



screw propellers. 321 

Questions and Problems. 
Define the following terms as applied to the screw propeller : 

1. Diameter. 6. Length of screw. 

2. Pitch. 7. Speed of screw. 

3. Disk area. 8. Pitch ratio. 

4. Developed area. 9. Slip ; and how expressed. 

5. Projected area. 10. Expanding pitch. 

In designing a screw propeller what are the three elements of 
greatest importance? 

Given the total I. H. P. (both main engines) =16,500; speed of 
ship = 18 knots; E. P. M. = 120; find the pitch and diameter of the 
screws (using Burean constants). Check by comparison with the 
following screw: I. H. P. (both main engines) =10,890; E. P. M. 
= 128.25; speed of ship = 16.79 knots; diameter of screw = 15 feet; 
pitch ratio = 1.03. Assume a slip of 14^. After finding diameter 
check b}^ the following: speed at periphery < 6500 feet per minute. 

Given total I. H. P. (both main engines) =16,000; speed of ship 
= 18^ knots; E. P. M. = 120; diameter of screw = 17 feet 9 inches 
pitch =17 feet 9 inches; find necessary developed area of each screw. 
Check by indicated thrust. Efficiency of engine = 90^ efficiency of 
screw = 62;^. 

Given the total I. H. P. (both main engines) =11,000; E. P. M. 
= 128 ; each screw has three blades. Find the maximum breadth of 
the screw and the thickness of the blade at shaft axis and at tip. 
Diameter of screw = 15 feet. 

Given a screw propeller 15 feet diameter and 16 feet pitch; three 
blades; diameter of pitch circle of blade studs = 25 inches; distance 
between blade flange and shaft axis = 14 inches; number of studs in 
each blade = 7; I. H. P. (one engine) =4500; E. P. M. = 128; Tobin 
bronze studs. Find the nominal diameter of each stud. 

A ship whose total I. H. P. (both main engines) is 9000; speed 

15 knots at 128 E. P. M., has twin screws each 15 feet diameter and 

16 feet pitch. What value of the constant K was used in determin- 
ing the diameter of the screws ? What is the fc slip ? 

A ship whose total I. H. P. (both engines) is 9000 at 128 E. P. M., 
has a total helicoidal area of loth screws of 132 square feet. What 
value of the constant K was used in determining the developed 
area? 



322 ENGINEERING MECHANICS. 

Explain how the surface of a screw propeller is generated, showing 
how the angle changes with the radius. Show the value of the 
angle of the helix. Explain the approximate method of finding the 
expanded blade surface. 

Make a sketch (plan, front and side elevations) of one blade of the 
following propeller: Helicoidal area (each blade) =28 square feet; 
diameter of screw = 16.75 feet; thickness of blade at shaft axis 
= 9J"; thickness of blade at tip = l"; diameter of hub = -J- diameter 
of screw ; pitch = 17.25 feet. Assume that J of the area of the ellipse 
falls within the hub. 

Make a sketch (plan, front and side elevations) of one blade of 
the following propeller: helicoidal area (each blade) =32 square 
feet ; diameter of hub = f of diameter of screw ; diameter of screw = 
18 feet; mean pitch=19 feet; thickness of blade (considered as 
produced) at shaft axis = 12"; thickness of blade at tip = l". As- 
sume that i of the area of the ellipse falls within the hub. 

Make a sketch (two views) of the hub for a three-bladed screw 
propeller, showing clearly the method of securing the hub to the 
shaft and of securing the blades to the hub. 

Make a sketch (plan, front and side elevations) of one blade of 
the following propeller: helicoidal area (each blade) =22 square 
feet; diameter of screw = 15 feet; mean pitch = 16 feet; diameter of 
hub=|-of diameter of screw; thickness of blade (considered as 
produced) at shaft axis = 7|"; thickness of blade at tip = f". As- 
sume that J of the area of the ellipse falls within the hub. 



USEFUL TABLES. 



USEFUL TABLES. 

Standard Dimensions of Bolts and Nuts for the United States Navt. 



Diameter. 


Area. 


Thr'ds. 


Long Diameter. 


Short D. 


Depth. 


Norn. 


Effect. 


Norn. 


Effect. 


No. 


Hex. 


Square. 


H 


.«feSq. 


Head. 


Nut. 


1/4 


.185 


.049 


.026 


20 


9/16 


23/32 




1/2 


1/4 


1/4 


5/16 


.240 


.077 


.045 


18 


11/16 


27/32 




19/32 


19/64 


5/16 


3/8 


.294 


.110 


.067 


16 


25/32 


31/32 




11/16 


11/32 


3/8 


7/1 C 


.345 


.150 


.093 


14 


29/32 


1 3/32 




25/32 


25/64 


7/16 


1/2 


.400 


.196 


.125 


13 


1 


1 1/4 




7/8 


7/16 


1/2 


9/16 


.454 


.249 


.162 


12 


1 1/8 


1 3/8 




31/32 


31/64 


9/16 


5/8 


.507 


.307 


.303 


11 


1 7/33 


1 1/2 




1/16 


17/33 


5/8 


3/4 


.620 


.443 


.303 


10 


1 7/10 


1 3/4 




1/4 


5/8 


3/4 


7/8 


.731 


.601 


.419 


9 


1 21/33 


2 1/32 




7/16 


23/32 


7/8 


1 


.837 


.785 


.550 


8 


1 7/8 


2 5/16 




5/8 


13/16 


1 


1 1/8 


.940 


.994 


.694 


7 


2 3/33 


2 9/16 




13/16 


29/32 


1 1/8 


1 1/4. 


1.065 


1.237 


.891 


7 


2 5/16 


2 27/32 


2 




1 


1 1/4 


1 3/8 


1.160 


1.4S5 


1.057 


6 


2 17/33 


3 3/32 


2 


3/16 


1 3/32 


1 3/8 


1 1/2 


1.2S4 


1.767 


1 . 294 


6 


2 3/4 


3 11/33 


2 


3/8 


1 3/16 


1 1/3 


1 5/8 


1.389 


2.074 


1.515 


5 1/3 


2 31/33 


3 5/8 


2 


9/16 


1 9/33 


1 5/8 


1 3/4 


1.491 


2 . 405 


1.746 


5 


3 3/16 


3 7/8 


2 


3/4 


1 3/8 


1 3/4 


1 7/8 


1.616 


2.761 


2.051 


5 


3 13/33 


4 5/32 


3 


15/16 


1 15/32 


1 7/8 


2 


1.712 


3.143 


2.303 


4 1/3 


3 19/33 


4 13/32 


3 


1/8 


1 9/16 


2 


2 1/4 


1.962 


3.976 


3.023 


4 1/3 


4 1/33 


4 15/16 


3 


1/3 


1 3/4 


2 1/4 


2 1/2 


2.176 


4.909 


3.719 


4 


4 15/33 


5 15/32 


3 


7/8 


1 15/16 


2 1/3 


2 3/4 


2.426 


5.940 


4.622 


4 


4 29/32 


6 


4 


1/4 


2 1/8 


2 3/4 


3 


2.676 


7.069 


5.624 


4 


5 11/32 


6 17/32 


4 


5/8 


2 5/16 


3 


3 1/4 


2.936 


8.296 


6.724 


4 


5 25/33 


7 1/16 


5 




2 1/2 


3 1/4 


3 1/2 


3.176 


9.621 


7.922 


4 


6 7/33 


7 19/32 


5 


3/8 


2 11/16 


3 1/3 


3 3/4 


3.426 


11.04 


9.219 


4 


6 5/8 


8 1/8 


5 


3/4 


2 7/8 


3 3/4 


4 


3.676 


12.57 


10.61 


4 


7 1/10 


8 21/32 


6 


1/8 


3 1/16 


4 


4 1/4 


3.926 


14.19 


12.11 


4 


7 1/3 


9 3/16 


6 


1/2 


3 1/4 


4 1/4 


4 1/2 


4.176 


15.90 


13.70 


4 


7 15/16 


9 23/32 


6 


7/8 


3 7/16 


4 1/3 


4 3/4 


4.426 


17.72 


15.39 


4 


8 3/8 


10 1/4 


7 


1/4 


3 5/8 


4 3/4 


5 


4.676 


19.64 


17.17 


4 


8 13/16 


10 25/32 


7 


5/8 


3 13/16 


5 


5 1/4 


4.926 


21.65 


19.06 


4 


9 1/4 


11 5/16 


8 




4 


5 1/4 


5 1/2 


5.176 


23.76 


21.04 


4 


9 11/16 


11 27/32 


8 


3/8 


4 3/16 


5 1/3 


5 3/4 


5.426 


25.97 


23.12 


4 


10 3/32 


13 3/8 


8 


3/4 


4 3/8 


5 3/4 


6 


5.676 


28.27 


25.30 


4 


10 17/32 


12 29/32 


9 


1/8 


4 9/16 


6 



II 



ENGINEERING MECHANICS. 



Decimals and Eractions of an Inch. 



1/33 


.031 


9/32 


.281 


17/32 


.531 


25/32 .781 


1/16 


.062 


5/16 


.312 


9/16 


.562 


13/16 .812 


3/32 


.094 


11/32 


.344 


19/32 


.594 


27/32 .844 


1/8 


.125 


3/8 


.375 


5/8 


.625 


7/8 .875 


5/32 


.156 


13/32 


.406 


21/32 


.656 


29/32 .906 


3/16 


.188 


7/16 


.438 


11/16 


.688 


15/16 .938 


7/32 


.219 


15/32 


.409 


23/32 


.719 


31/32 .969 


1/4 


.250 


1/2 


.500 


3/4 


.750 


1 1.000 



Table of Areas of Circles, advancing by Eighths. 



Diam. 




+V' 


+V' 


+%" 


+W 


+%" 


+%" 


+V' 


0" 


.000 


.012 


.049 


.110 


.196 


.307 


.441 


.601 


1'' 


.785 


.994 


1.227 


1.485 


1.767 


2.G74 


2.405 


2.761 


2" 


3.142 


3.547 


3.976 


4.430 


4.909 


5.412 


5.940 


6.492 


3" 


7.069 


7.670 


8.296 


8.946 


9.621 


10.32 


11.05 


11.79 


4" 


12.57 


13.36 


14.19 


15.03 


15.90 


16.80 


17.73 


18.67 


5" 


19.64 


20.63 


21.65 


22.69 


23.76 


24.85 


25.97 


27.11 


6'' 


28.27 


29.46 


30.68 


31.92 


33.18 


34.47 


35.73 


37.12 


V 


88.48 


39.87 


41.28 


43.72 


44.18 


45.66 


47.17 


48.71 


8" 


50.27 


51.85 


53.46 


55.09 


56.74 


58.43 


60.13 


61.86 


9'' 


63.62 


65.40 


67.20 


69.03 


70.88 


72.76 


74.66 


76.59 


10'/ 


78.54 


80.52 


82.52 


84.54 


86.59 


88.66 


90.76 


92.89 



^2 

Values of — , when d advances by Eighths. 
16 



d 




+V 


+V' 


+%" 


+V 


+%" 


^%" 


+%" 


0" 


.0000 


.0010 


.0039 


.0088 


.0156 


.0344 


.0353 


.0479 


1" 


.0625 


.0791 


.0977 


.1183 


.1406 


.1650 


.1914 


.2197 


2" 


.2500 


.2822 


.3164 


.3536 


.3906 


.4307 


.4727 


.5166 


3'' 


.5625 


.6104 


.6596 


.7119 


.7656 


.8313 


.8789 


.9385 


4" 


1.000 


1.064 


1.129 


1.196 


1.266 


1.337 


1.410 


1.485 


5'' 


1.563 


1.643 


1.723 


1.806 


1.891 


1.978 


2.066 


2.157 


6'' 


2.250 


2.345 


2.441 


3.540 


2.641 


2.743 


2.848 


2.954 


1" 


3.063 


3.173 


3.285 


3.399 


3.516 


3 . 634 


3.754 


3.876 


%" 


4.000 


4.136 


4.354 


4.384 


4.516 


4.649 


4.785 


4.933 


9" 


5.063 


5.204 


5.348 


5.493 


5.641 


5.777 


5.941 


6.095 


10'' 


6.250 


6.407 


6.566 


0.737 


6.891 


7.056 


7.332 


7.391 



TRIGONOMETRIC FORMULiE. Ill 



Trigonometric Formula. 

sin (x + y) =sinx cos y + cos x siny. 

cos (x + y) = cos X cosy — sin X siny. 

sin (x — y) =sin x cos y — cos x sin y. 

cos (x — y ) = cos X cos y + sin x sin y. 

2 sin X cos y = sin (x + y) +sin (x — y). 

2 cosx sin y=sin (x + y) —sin (x — y). 

2 cos X cosy = cos (x + y) +cos (x — y). 

2 sinx sin y = cos (x — y) —cos (x + y). 

sin x + siny = 2 sin J (^ + y) cos^ (x — y). 

sin X — siny = 2 cos -J (x + y) sin -J (x — y). 

cos y — cos x = 2 sin -J (x + y) sin|^ (x — y). 

cos x + cos y = 2 cos -J (x + y) cos -J (x — y). 

sin X — sin V _ tan-| (x + y) 
sinx — siny tan-|(x — y)* 

, / , \ tan X + tan y 
tan (x + y) = — — — ^-- — ^ . 
1 — tan X tany 

tan(x-y)= ^^^^^-^f^y . 
^ -^ ^ 1 + tanxtany 

, / , X cot X cot y — 1 

cot (x + y) = ^- — . 

^ ^ ^ cotx + coty 

, / s cot X cot y + 1 

cot (x — y) = — ~ ^—r — . 

^ "^ cot X — cot y 

, X ,t;oN tanxH^l 
tan (x±45 ) 



tan X ± 1 

sin (x + y) sin (x — y) =sin^ x— sin^ y^cos^ y — cos^ x. 
cos (x + y) cos (x — y) =:cos^ X— sin^y=cos^y — sin^x. 

tan X ± tan y = ^ ~-^' . 

cos X cos y 

sm X sm y 

2tanx 



sin 2x = 2 sin x cos x : 



1 + tan- X 

1 — tan^ X 



cos 2x = cos^ X — sin^ x==l — 2 sin^ x=2 cos^ x — 1 = 



1 + tan^ X 



IV ENGINEERING MECHANICS. 

1 — cos2x 2 sin^ X , , 



l + cos2x 


2 eos^ X 


, ^ Q 2 tan X 
^^^^^=l-tan^x- 


cot2x=^f ^-1. 
2cotx 


sin3x = 3 


sinx — 4sin^x. 


COB 3X=:4 


cos^x — 3 cosx. 


, „ 3 tanx— tan^ X 

tan 3x= -^ — ^r- — o • 

1 — 3 tan^ X 


^^^(1-) 


1 — cosx 

2 


cos^( 2 ) 


1 + cosx 

2 


tan-^ X + 


tan-^v-tan-if .^+7 



Integrals. 



x"(ix= ^— - 
n + 1 

dx 1 
—— = logx. 



a^dx= ^ 



log a * 
cos M^=sin 0. 

sin ^d^=: —cos B. 

sec^M^rztan^. 

cosec^ ^d^= —cot 0, 

sec (9 tan M^r=sec ^. 

cosec 6 cot ^d^= — cosec 0, 



dx 

^ = sin"^ X + c == — cos"^ X + c'. 



VI 



III 



INTEGRALS. 



h c = — COS"^ h c'. 

a a 



taii~^x + c = — cot~^ X + c'. 



tan-^ h c = cot^ he/ 



+x- a a a a 

r dx 

I / ^ ? - = sec"^ X + c = — cosec"^ x + c . 
JxVx^ — 1 

[ dx 1 , X ■ 1 , X , , 

jxVx- — a^ a a a a 

[ dx 

-7==- =: vers ^ X. 
J V 2x — X- 



dx , X 

^ =vers~^ — 



V2ax- 

tan xdx= —log cos x 

taii2^d^=taii(9-^. 



f 

udv = iiv — vdu. 

f dx _ J , / x— a 

J(x-a)(x-b) - a-b ^°^lx-b 

r dx 1 , /x— a\ 

Jx^:^^= 2^^°gU-bj• 
j'siIl2 M6'=i(6'-siii ^ cos 0). 

fcos^ ^d^ = ^(^ + sin ^ cos 6). 



loo" tan ^. 



Ism V cos 

1 — cos 



sin ^ 



1 , r Vb + a+Vb-a(tan^ ^)-i 
- Vb^-a^ ^^LVb + a:-Vb=^(tani^)J' 



VI ENGINEERING MECHANICS. 

dx 



cos*" OdiO (when m is positive and even) 



sin 



_ (m-l)(ni-3) .... l ^-\iimit 
m(m-2) .... 2 Jumit 

(when m is positive and odd) 

m(m — 2) .... 1 -iiimit 

f Bin". COS" Mg= <"^-l)("'-f)/---(%-:^Hn-3).. 

Jo (m + n)(ni + n-2) .... ' 

where a=: 1, unless m and n are both even, in which case a=-^ , 




^ 





aSJXSRAr. AnnATvaKnf^ ' T of MAcrnTVKnT. 



s it ^- 




■V. 






s ' ■«' r 













\ 







0<>^ 



•nt V* 












0^^. 



<^ ^ « w 



7 -f. 






■"oo^ 



0' 









v^ 



3C3 






< 



■^o. 






X^^^. 



s o ^ *C> 



^- 



\V^ , , . , *>- * ., s ' ,f° 



>0' 






V- 



, X -^ .A 



- ^^ 



^r^.-'-o 




'^oo^ 
















'% 






^^.o 



^0, X 






.K:- 









°- '.try/ .^^ -% 



*^ 



^°^ 



